Sequences and series

Class 11 Mathematics, unit 4, book pages 116–149

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A sequence is an ordered list of numbers, and a series is the sum of the terms of a sequence. This unit studies the three standard progressions: arithmetic (add the same number each time), geometric (multiply by the same number each time) and harmonic (the reciprocals form an arithmetic sequence). For each it finds the $n$ th term, the means between two numbers and the sum, including the sum of an infinite geometric series. It then introduces sigma notation, the sums of the first $n$ natural numbers, their squares and their cubes, and the arithmetico-geometric series, and finishes with problems on saving, population and infection.

Learning outcomes (9)

Solve problems by analysing arithmetic sequence and series up to $n$ terms.
Solve problems by analysing geometric sequence and series up to $n$ terms.
Identify a sequence as arithmetic or geometric sequence up to $n$ terms.
Solve problems by analysing harmonic sequences and series up to $n$ terms.
Recognize sigma ( $Σ$ ) notation.
Find sum of the first $n$ natural numbers ( $Σ n$ ), the squares of the first $n$ natural numbers ( $Σ n^{2}$ ) and the cubes of the first $n$ natural numbers ( $Σ n^{3}$ ).
Define arithmetico-geometric series.
Find sum to $n$ terms of the arithmetico-geometric series.
Solve real life problems involving arithmetic sequence, arithmetic mean and arithmetic series.

4.1Sequence

We meet sequences at the very beginning of our mathematical experience. The list of even numbers $2, 4, 6, 8, 10, \dots$ and the list of odd numbers $1, 3, 5, 7, 9, \dots$ are examples. We can 'predict' the 20th term of each list just by using common sense. A sequence is simply an ordered list of numbers.

Finite and infinite sequences

A sequence is either finite or infinite.

Sequence	Kind	Reason
$2, 4, 6, 8, 10$	finite	it has exactly five terms
$5, 10, 15, 20, 25, \dots$	infinite	it continues without bound
$1, 3, 5, 7, 9, \dots$ (positive odd numbers)	infinite	there is no last term
$1, 3, 5, 7, 9, \dots, 99$ (positive odd numbers less than $100$ )	finite	the last term is $99$ and there are $50$ terms

We use the symbol $a_{n}$ for the $n$ th term of a given sequence. For the positive odd numbers $a_{1} = 1$ , $a_{2} = 3$ , $a_{3} = 5$ and so on. The first term is $a_{1} = 1$ , but there is no last term.

Ways to display a sequence

Write out the first few terms.
Give a formula for the general term.
Give a recurrence relation, that is, a rule that produces each term from the one before it.

A sequence as a function

If the domain of a function $f$ is the set of positive integers, the elements of its range can be arranged in order of increasing $n$ :

f (1), f (2), f (3), \dots, f (n), \dots

Functions whose domain is the whole set of positive integers are given a special name.

Sequence: A sequence is a function whose domain is the set of positive integers.

Instead of the usual function notation $f (x)$ , a sequence is written with the symbol ${a_{n}}$ . The terms are formed by letting $n$ take the values $1, 2, 3, \dots$ in the general term $a_{n}$ . So ${a_{n}}$ is the list $a_{1}, a_{2}, a_{3}, \dots, a_{n}, \dots$ in which each element of the domain is matched with an element of the range.

The domain and range of a sequence. Each positive integer

n

(below) is sent by the function to exactly one term

a_{n}

(above).

The general term

A much better way to describe a sequence is to give a formula for the $n$ th term $a_{n}$ . This is called the general term of the sequence. For the sequence of odd numbers, $a_{n} = 2 n - 1$ is the general term.

Take the sequence $2, 4, 8, 16, \dots$ . The first term is $a_{1} = 2^{1} = 2$ , the second term is $a_{2} = 2^{2} = 4$ and the third term is $a_{3} = 2^{3} = 8$ . So the general term is $a_{n} = 2^{n}$ , and the sequence can be written $2, 4, 8, \dots, 2^{n}, \dots$

Example 1

Find the first four terms and the 57th term of the sequence whose general term is

a_{n} = \frac{( - 1 ) ^{n}}{n + 1}

Put $n = 1, 2, 3, 4$ in turn into the general term.
$a_{1} = \frac{( - 1 ) ^{1}}{1 + 1} = - \frac{1}{2}, a_{2} = \frac{( - 1 ) ^{2}}{2 + 1} = \frac{1}{3}$
Continue with $n = 3$ and $n = 4$ .
$a_{3} = \frac{( - 1 ) ^{3}}{3 + 1} = - \frac{1}{4}, a_{4} = \frac{( - 1 ) ^{4}}{4 + 1} = \frac{1}{5}$
For the 57th term put $n = 57$ . The power $57$ is odd, so $(- 1)^{57} = - 1$ .
$a_{57} = \frac{( - 1 ) ^{57}}{57 + 1} = - \frac{1}{58}$

- \frac{1}{2}, \frac{1}{3}, - \frac{1}{4}, \frac{1}{5}; a_{57} = - \frac{1}{58}

Example 2

For each sequence, predict the general term: (i)

1, 4, 9, 16, 25, \dots

(ii)

1, 2, 3, 4, \dots

(iii)

- 1, 2, - 4, 8, - 16, \dots

(iv)

2, 4, 8, 16, \dots

(i) These are the squares of consecutive positive integers $1^{2}, 2^{2}, 3^{2}, \dots$ , so the $n$ th term is the square of $n$ .
$a_{n} = n^{2}$
(ii) These are the square roots of consecutive positive integers.
$a_{n} = n$
(iii) The numbers $1, 2, 4, 8, 16$ are powers of $2$ starting from $2^{0}$ , so the size of the $n$ th term is $2^{n - 1}$ . The signs alternate, starting negative, which is the job of $(- 1)^{n}$ .
$a_{n} = (- 1)^{n} 2^{n - 1}$
(iv) The pattern of powers of $2$ gives $16$ as the next term after $8$ , so each term is $2$ raised to its position.
$a_{n} = 2^{n}$

4.2, 4.2.1Arithmetic sequenceOften examined

A professional race car driver drives out of a curve. He enters the straight at $119.9 mph$ . He increases his speed by $78.3 mph$ and after $9$ seconds his speed is $198.2 mph$ . The table shows how his speed increased each second after entering the straight.

Speed of the race car in each second after it enters the straight.
Number of seconds	0	1	2	3	4	5	6	7	8	9
Speed in mph	119.9	128.6	137.3	146.0	154.7	163.4	172.1	180.8	189.5	198.2

Speed of the race car in each second after it enters the straight.
Number of seconds	Speed in mph
0	119.9
1	128.6
2	137.3
3	146.0
4	154.7
5	163.4
6	172.1
7	180.8
8	189.5
9	198.2

Speed of the race car

100

150

200

250

time (s)

speed (mph)

(0, 119.9)

(9, 198.2)

The ten speeds lie on a straight line, because the same

8.7 mph

is added every second. A sequence with a common difference always plots as points on a line.

From the table and graph we observe a number pattern. This set of numbers is an example of a sequence. Each number in a sequence is called a term. The first term is written $a_{1}$ , the second term $a_{2}$ , and so on up to $a_{n}$ , the $n$ th term. The sequence in the table contains ten terms: $a_{1} = 119.9$ , $a_{2} = 128.6$ and $a_{10} = 198.2$ .

Each term is obtained by adding $8.7$ to the previous term. A sequence of this type is called an arithmetic sequence or arithmetic progression (A.P.). The number added to find the next term is called the common difference and is written $d$ .

Arithmetic sequence (A.P.): A sequence in which every term after the first is found by adding the same number to the preceding term.
Common difference $d$: The number that is added each time. It is the difference between any term and the one before it: $d = a_{n + 1} - a_{n}$ .

To find the next terms of an arithmetic sequence, first find the common difference $d$ by subtracting any term from the term that follows it. Then add the common difference to the last term to find the successive terms.

Example 3

Find the next four terms of the arithmetic sequence

33, 39, 45, \dots

Find the common difference by subtracting two consecutive terms. Either pair gives the same answer.
$d = 39 - 33 = 6 or d = 45 - 39 = 6$
Add $6$ to the last term and keep adding until four new terms are found.
$a_{4} a_{5} a_{6} a_{7} = 45 + 6 = 51 = 51 + 6 = 57 = 57 + 6 = 63 = 63 + 6 = 69$

51, 57, 63, 69

Formula for the nth term of an arithmetic sequence

Adding $d$ to a term gives the next one, so the terms can be built up from $a_{1}$ one step at a time.

Derivation The nth term of an A.P.

Aim. Show that the $n$ th term of an arithmetic sequence with first term $a_{1}$ and common difference $d$ is $a_{n} = a_{1} + (n - 1) d$ .

Start with the first term and add $d$ each time.
$a_{1} = a_{1}, a_{2} = a_{1} + d$
The third term is the second plus $d$ , so it contains $d$ twice. The fourth contains $d$ three times.
$a_{3} = a_{2} + d = a_{1} + 2 d, a_{4} = a_{3} + d = a_{1} + 3 d$
In every case the number of $d$ 's is one less than the subscript of the term. So the $n$ th term contains $(n - 1)$ copies of $d$ .
$a_{n} = a_{1} + (n - 1) d$

Result

a_{n} = a_{1} + (n - 1) d

The coefficient of $d$ is always $1$ less than the subscript. Given $a_{1}$ and $d$ you can find any term directly, without listing the earlier ones.

Example 4

Suppose a race car driver increases speed at a constant rate. What will his speed be after

15

seconds if his initial speed is

85 mph

and his rate of acceleration is

4.5 mph

per second?

The initial speed is the first term, and $4.5$ is added every second. 'After $15$ seconds' means we need the $16$ th term, because the first term is at second $0$ .
$a_{1} = 85, d = 4.5, a_{16} = ?$
Use $a_{n} = a_{1} + (n - 1) d$ with $n = 16$ .
$a_{16} = 85 + (16 - 1) (4.5) = 85 + 67.5 = 152.5$

152.5 mph

Example 5

The third term of an arithmetic sequence is

8

and the sixteenth term is

47

. Find

a_{1}

and

d

and construct the sequence. Also find

a_{15}

Write each given term with the formula $a_{n} = a_{1} + (n - 1) d$ . For $a_{3} = 8$ we have $n = 3$ .
$8 = a_{1} + 2 d (i)$
For $a_{16} = 47$ we have $n = 16$ .
$47 = a_{1} + 15 d (ii)$
Subtract (i) from (ii) to remove $a_{1}$ , then find $d$ and $a_{1}$ .
$39 = 13 d ⟹ d = 3, a_{1} = 8 - 2 (3) = 2$
Build the sequence by adding $3$ each time.
$a_{2} = 2 + 3 = 5, a_{3} = 2 + 6 = 8, a_{4} = 2 + 9 = 11$
Now find $a_{15}$ with $n = 15$ .
$a_{15} = a_{1} + 14 d = 2 + 14 (3) = 44$

a_{1} = 2, d = 3; 2, 5, 8, 11, \dots; a_{15} = 44

Worked example: an A.P. from two terms, and is a number a term?

In an A.P. the 4th term is

17

and the 9th term is

42

. Find

a_{1}

d

and

a_{n}

. Is

47

a term? Is

300

a term?

The terms lie on a line

n

a_{n}

(1, 2)

(4, 17)

(9, 42)

(10, 47)

The terms

a_{n} = 5 n - 3

are the points of the line at whole

n

. The line passes through

(4, 17)

and

(9, 42)

; only whole values of

n

are terms.

Write both terms with $a_{n} = a_{1} + (n - 1) d$ .
$a_{1} + 3 d = 17, a_{1} + 8 d = 42$
Subtract to remove $a_{1}$ : the five steps from the 4th to the 9th term add $42 - 17 = 25$ .
$5 d = 25 ⟹ d = 5$
Put $d = 5$ into the first equation.
$a_{1} = 17 - 15 = 2$
The general term.
$a_{n} = 2 + (n - 1) 5 = 5 n - 3$
$47$ is a term only if $n$ comes out a positive whole number.
$5 n - 3 = 47 ⟹ n = 10 so 47 is the 10th term$
Same test for $300$ .
$5 n - 3 = 300 ⟹ n = \frac{303}{5} = 60.6 not a whole number, so 300 is not a term$

a_{1} = 2, d = 5, a_{n} = 5 n - 3; 47 is the 10th term; 300 is not a term

4.2.2Arithmetic meanOften examined

To find the arithmetic mean between two numbers a and b we use the formula A.M.=\dfrac{a+b}{2}. The reason is the definition below.

Arithmetic mean (A.M.): A number A is the arithmetic mean between two numbers a and b if a,\,A,\,b are in A.P.

Derivation The arithmetic mean of two numbers

Aim. Find A when a,A,b are in A.P.

In an A.P. the common difference is the same between each pair of neighbouring terms.
d=A-a=b-A
Clear the equation A-a=b-A by collecting A on one side.
2A=a+b
Divide by 2.
A=\dfrac{a+b}{2}

ResultA=\dfrac{a+b}{2}

The arithmetic mean is the ordinary average of the two numbers. It sits exactly half way between a and b.

Example 6

Find the four arithmetic means between 19 and 54.

Write the sequence with the four unknown means in the gaps: 19,\_,\_,\_,\_,54. There are six terms, so a_1=19 and a_6=54. Use the nth term formula to find d.
a_6=a_1+5d\;\Longrightarrow\;54=19+5d\;\Longrightarrow\;d=7
Now add d=7 repeatedly to find the means.
\begin{aligned}a_2&=19+7=26\\ a_3&=19+2(7)=33\\ a_4&=19+3(7)=40\\ a_5&=19+4(7)=47\end{aligned}

26,\ 33,\ 40,\ 47

Example 7

Find the A.M. between 6 and 18.

Put a=6 and b=18 into the formula.
A.M.=\dfrac{a+b}{2}=\dfrac{6+18}{2}=12

Example 8

Find the 7 A.Ms between 7 and 20.

Let A_1,A_2,\dots,A_7 be the required means. Then 7,A_1,A_2,\dots,A_7,20 are in A.P. There are nine terms, so a_1=7 and a_9=20.
Use a_9=a_1+8d to find the common difference.
7+8d=20\;\Longrightarrow\;8d=13\;\Longrightarrow\;d=\dfrac{13}{8}
Add d repeatedly: A_k=7+kd.
A_1=7+\dfrac{13}{8}=\dfrac{69}{8},\quad A_2=7+\dfrac{26}{8}=\dfrac{41}{4},\quad A_3=7+\dfrac{39}{8}=\dfrac{95}{8}
Similarly for the rest.
\begin{aligned}A_4&=7+\dfrac{52}{8}=\dfrac{27}{2}\\ A_5&=\dfrac{121}{8}\\ A_6&=7+\dfrac{78}{8}=\dfrac{67}{4}\\ A_7&=7+\dfrac{91}{8}=\dfrac{147}{8}\end{aligned}
Check: one more step from A_7 must give 20.
A_7+d=\dfrac{147}{8}+\dfrac{13}{8}=\dfrac{160}{8}=20

A_1=\dfrac{69}{8},\ A_2=\dfrac{41}{4},\ A_3=\dfrac{95}{8},\ A_4=\dfrac{27}{2},\ A_5=\dfrac{121}{8},\ A_6=\dfrac{67}{4},\ A_7=\dfrac{147}{8}

Check point

The book asks you to show that the sum of n arithmetic means between a and b is equal to n times the single arithmetic mean between a and b.

Derivation Check point: the sum of n arithmetic means

Aim. Show that if A_1,A_2,\dots,A_n are n A.Ms between a and b, then A_1+A_2+\dots+A_n=n\cdot\dfrac{a+b}{2}.

The sequence a,A_1,\dots,A_n,b has n+2 terms, so b=a+(n+1)d. This gives the common difference.
d=\dfrac{b-a}{n+1}
The kth mean is A_k=a+kd. Pair the kth mean from the start with the kth mean from the end.
A_k+A_{n+1-k}=2a+(n+1)d=2a+(b-a)=a+b
Let \Sigma=A_1+A_2+\dots+A_n. Write it forwards and backwards and add the two lines. Each of the n column sums equals a+b.
2\Sigma=n(a+b)
Divide by 2.
\Sigma=n\cdot\dfrac{a+b}{2}

ResultA_1+A_2+\dots+A_n=n\cdot\dfrac{a+b}{2}

The n means together add up to n times the one mean \dfrac{a+b}{2}. In Example 8 the seven means add to 7\times13.5=94.5, and indeed \dfrac{69+82+95+108+121+134+147}{8}=\dfrac{756}{8}=94.5.

Challenge

Challenge (book box)

Find three numbers that have a sum of 27, a product of 288 and form an arithmetic sequence.

Three numbers in A.P. can be written with the middle number a and common difference d as a-d,\ a,\ a+d. Their sum removes d.
(a-d)+a+(a+d)=3a=27\;\Longrightarrow\;a=9
Use the product =288. The product of the outer two is (a-d)(a+d)=a^2-d^2.
9\left(81-d^2\right)=288
Divide by 9 and solve for d.
81-d^2=32\;\Longrightarrow\;d^2=49\;\Longrightarrow\;d=\pm7
Take d=7 (the choice d=-7 gives the same numbers in reverse order).
9-7=2,\quad 9,\quad 9+7=16
Check the sum and product.
2+9+16=27,\qquad 2\times9\times16=288

2,\ 9,\ 16

4.3, 4.3.1Arithmetic seriesOften examined

A sky diver falls freely covering the distances 16,\,48,\,80,\,112,\,144,\,176,\dots in successive seconds (in feet). These free-fall distances form an arithmetic sequence. To find the total distance covered by the sky diver we would add the terms of the sequence:

16+48+80+112+144+176

The indicated sum of the terms of a sequence is called a series. The series above is called an arithmetic series.

Arithmetic series: The sum of the terms of an arithmetic sequence. An arithmetic sequence is a sequence of numbers in which the difference between consecutive terms is constant.

Here are some arithmetic sequences and their corresponding arithmetic series.

Arithmetic sequence	Arithmetic series
2,4,6,8,10	2+4+6+8+10
-8,-2,4	-8+(-2)+4
\dfrac45,\dfrac85,\dfrac{12}5,\dfrac{16}5	\dfrac45+\dfrac85+\dfrac{12}5+\dfrac{16}5
a_1,a_2,a_3,a_4,\dots,a_n	a_1+a_2+a_3+a_4+\dots+a_n

The symbol S_n stands for the sum of the first n terms of a series. For example, S_4 means the sum of the first four terms. The sum of the series 3+6+9+12 is S_4=30. If a series has many terms it is not convenient to list all of them and then add, so we look for a formula.

Sum of the nth term of an A.P. (sum of n terms)

To develop a formula for the sum of any arithmetic series, start with the sky diver series.

The sky diver example

Write S_6 in two different orders, forwards and backwards.
S_6=16+48+80+112+144+176
Write it again with the terms reversed and add the two lines column by column. Every column adds to the same number, 192.
\begin{aligned} S_6&=16+48+80+112+144+176\\ +\;S_6&=176+144+112+80+48+16\\ \hline 2S_6&=192+192+192+192+192+192\end{aligned}
There are 6 terms, so there are six sums of 192. Divide each side by 2.
2S_6=6\,[192]\;\Longrightarrow\;S_6=\dfrac62\,[192]=576

Here 6 is the number of terms n, and 192 is the sum of the first and last terms, a_1+a_n. Now do the same for any arithmetic series.

Derivation The sum of the first n terms of an arithmetic series

Aim. Prove S_n=\dfrac n2\left[a_1+a_n\right] and S_n=\dfrac n2\left[2a_1+(n-1)d\right].

Write the series forwards, then backwards (starting from a_n and subtracting d each time).
\begin{aligned} S_n&=a_1+(a_1+d)+\dots+(a_n-d)+a_n\\ S_n&=a_n+(a_n-d)+\dots+(a_1+d)+a_1\end{aligned}
Add the two lines. In every column the d cancels, so each column adds to a_1+a_n. There are n columns.
2S_n=n\,(a_1+a_n)
Divide by 2.
S_n=\dfrac n2\,[a_1+a_n]\qquad\text{(i)}
We know a_n=a_1+(n-1)d. Replace a_n in (i) with this.
S_n=\dfrac n2\,[a_1+(a_1+(n-1)d)]
Combine the two a_1 terms.
S_n=\dfrac n2\,[2a_1+(n-1)d]\qquad\text{(ii)}

ResultS_n=\dfrac n2\,[a_1+a_n]=\dfrac n2\,[2a_1+(n-1)d]

Use (i) when the first and last terms are known. Use (ii) when the first term and the common difference are known.

Example 9

Find the sum of the first 100 positive integers.

First method. The series is 1+2+3+\dots+100 with the first and last terms known.

Here a_1=1 and a_n=a_{100}=100. Use S_n=\dfrac n2[a_1+a_n].
S_{100}=\dfrac{100}{2}\,[1+100]=50\times101=5050

Second method. The same series, using the common difference.

Here a_1=1, d=1 and n=100. Use S_n=\dfrac n2[2a_1+(n-1)d].
S_{100}=\dfrac{100}{2}\,[2(1)+(100-1)(1)]=50\,[101]=5050

5050

Example 10

Find the sum of the first 50 terms of an arithmetic series where a_1=5 and d=25.

The given values are a_1=5, d=25, n=50. The last term is not given, so use the second formula.
S_n=\dfrac n2\,[2a_1+(n-1)d]
Substitute the values.
S_{50}=\dfrac{50}{2}\,[2(5)+(50-1)(25)]
Simplify inside the bracket: 49\times25=1225.
S_{50}=25\,[10+1225]=25\times1235=30875

30875

Example 11

Theatres are often built with more seats per row as the rows move towards the back. Suppose the main floor of a theatre has 28 seats in the first row, 32 in the second, 36 in the third and so on for 50 rows. How many seats are on the main floor?

The numbers of seats form an arithmetic sequence, so the total is an arithmetic series.
28+32+36+\dots\;(50\text{ rows})
Read off a_1=28, d=32-28=4 and n=50. Use S_n=\dfrac n2[2a_1+(n-1)d].
S_{50}=\dfrac{50}{2}\,[2(28)+(50-1)(4)]
Simplify.
S_{50}=25\,[56+196]=25\times252=6300

6300\text{ seats}

Example 12

Find the first three terms of an arithmetic series where a_1=17, a_n=101 and S_n=472.

First find n. We know a_1, a_n and S_n, so use S_n=\dfrac n2[a_1+a_n].
472=\dfrac n2\,[17+101]\;\Longrightarrow\;944=118\,n\;\Longrightarrow\;n=8
Next find d from a_n=a_1+(n-1)d with n=8.
101=17+(8-1)d\;\Longrightarrow\;84=7d\;\Longrightarrow\;d=12
Now find the first three terms.
a_2=a_1+d=17+12=29,\qquad a_3=a_1+2d=17+2(12)=41

17,\ 29,\ 41

Worked example: finding the number of terms from the sum

How many terms of the A.P. 5+9+13+\dots must be taken to make the sum 275?

Here a_1=5 and d=4. Put S_n=275 in S_n=\dfrac n2[2a_1+(n-1)d].
275=\frac n2[10+4(n-1)]=\frac n2(4n+6)=n(2n+3)
Rearrange into a quadratic equation in n.
2n^2+3n-275=0
Factorise the quadratic: look for two numbers with product 2\times(-275)=-550 and sum 3, namely 25 and -22. Split the middle term and factor.
2n^2+25n-22n-275=0\;\Longrightarrow\;n(2n+25)-11(2n+25)=0\;\Longrightarrow\;(n-11)(2n+25)=0
So n=11 or n=-\tfrac{25}{2}.
n=11\ \text{or}\ n=-\dfrac{25}{2}
A number of terms must be a positive whole number, so reject the negative root.
Check: a_{11}=5+10(4)=45, so S_{11}=\dfrac{11}{2}(5+45)=275.

n=11

4.4, 4.4.1Geometric sequenceOften examined

Iodine is used medically as a tracer isotope in monitoring the activity of the thyroid gland. A patient is given a compound containing radioactive iodine, and the amount retained by the gland is a measure of its ability to function. Iodine has a half-life of about 8 days. That means approximately every 8 days, half the mass of iodine decays into another element. Then in the next 8 days, half of the remaining iodine decays, and so on.

Suppose a container holds 64 milligrams of iodine. The remaining mass after each half-life is 64,\,32,\,16,\,8,\,4,\,2,\,1 and 0.5. This pattern of masses forms a sequence known as a geometric sequence or geometric progression (G.P.).

Mass of iodine after each half-life

The mass is halved each time, so the points 64,32,16,\dots,0.5 fall away along a curve. A geometric sequence with 0<r<1 decays, unlike an arithmetic sequence, whose points lie on a straight line.

Geometric sequence (G.P.): A sequence in which each term after the first is found by multiplying the previous term by the same number.
Common ratio r: The constant multiplier. It is found by dividing any term by the previous term: r=\dfrac{a_{n+1}}{a_n}. In the iodine sequence r=\dfrac12.

Example 13

Find the next two terms of the geometric sequence 4,12,36,\dots

Find the common ratio as the quotient of two consecutive terms.
r=\dfrac{12}{4}=\dfrac{36}{12}=3
Multiply the last term by r=3 to get the fourth, then the fifth term.
a_4=36(3)=108,\qquad a_5=108(3)=324

108,\ 324

Formula for the nth term of a geometric sequence

Successive terms of a geometric sequence are found by multiplying the previous term by r. So a geometric sequence is also a recursive sequence, and each term can be written as a product of the first term and a power of r. The table below uses Example 13 (a_1=4, r=3) and then the general case.

	a_1	a_2	a_3	a_4	\dots	a_n
Example 13	4	4(3)=12	4(3^2)=36	4(3^3)=108	\dots	4(3^{n-1})
General	a	ar	ar^2	ar^3	\dots	ar^{n-1}

Derivation The nth term of a G.P.

Aim. Show that the nth term of a geometric sequence with first term a_1 and common ratio r is a_n=a_1r^{n-1}.

Multiply by r each time we move one term along.
a_2=a_1r,\qquad a_3=a_2r=a_1r^2,\qquad a_4=a_3r=a_1r^3
The power of r is one less than the subscript of the term, as in the table. So for the nth term the power is n-1.
a_n=a_1r^{n-1}

Resulta_n=a_1\,r^{\,n-1}

Compare with the A.P. rule a_n=a_1+(n-1)d: in a G.P. we multiply by r a total of n-1 times, where in an A.P. we add d a total of n-1 times.

Example 14

Write the first five terms of a geometric sequence in which a_1=5 and r=2.

Use a_n=a_1r^{n-1} with a_1=5, r=2 and n=2,3,4,5.
a_2=a_1r=5(2)=10,\qquad a_3=a_1r^2=5(2)^2=20
Continue.
a_4=a_1r^3=5(2)^3=40,\qquad a_5=a_1r^4=5(2)^4=80

5,\ 10,\ 20,\ 40,\ 80

Example 15

Find the seventh term a_7 of a geometric sequence in which a_3=96 and r=4.

The third term is a_1r^{3-1}=a_1r^2. We need a_1 before we can find a_7.
96=a_1(4)^2=16\,a_1\;\Longrightarrow\;a_1=6
Now use a_7=a_1r^{6}.
a_7=6\,(4)^6=6\times4096=24576

24576

Example 16

Mr. Khalid saves Rs. 1000 on the first day. Each day after that he saves double the amount he saved the day before. Find the amount he should save on the 20th day of the month.

The amount is doubled each day, so the amounts form a geometric sequence with a_1=1000 and r=2.
a_{20}=a_1r^{19}
Substitute and use 2^{19}=524288.
a_{20}=1000\,(2)^{19}=1000\times524288=524288000

\text{Rs. }524{,}288{,}000

Worked example: A.P. and G.P. that start the same way

The sequences 2,4,6,8,\dots and 2,4,8,16,\dots both begin 2,4. Decide which is an A.P. and which a G.P., find the 6th and 8th terms of each, and compare.

A.P. against G.P.

Both sequences start 2,4. The A.P. rises by 2 each step and stays on a line; the G.P. doubles each step and climbs away from it.

Test the differences and the ratios of consecutive terms.
\begin{aligned}2,4,6,8:&\ 4-2=6-4=8-6=2\ (\text{A.P., }d=2)\\ 2,4,8,16:&\ \tfrac42=\tfrac84=\tfrac{16}8=2\ (\text{G.P., }r=2)\end{aligned}
A.P. terms: a_n=a_1+(n-1)d.
a_n=2+(n-1)2=2n:\quad a_6=12,\quad a_8=16
G.P. terms: a_n=a_1r^{n-1}.
a_n=2(2)^{n-1}=2^n:\quad a_6=2^6=64,\quad a_8=2^8=256
Compare. The A.P. adds the same amount each time, so its graph is a straight line; the G.P. multiplies by the same amount, so each step is larger than the last. Two sequences cannot be identified from their first two terms alone: the third term decides.

\text{A.P.: }a_6=12,\ a_8=16;\qquad\text{G.P.: }a_6=64,\ a_8=256

4.5Geometric mean

Geometric mean (G.M.): If a,\,G,\,b are in a geometric sequence, then G is called the geometric mean of a and b.

Derivation The geometric mean of two numbers

Aim. Find G when a,G,b are in G.P.

In a G.P. the ratio of each term to the one before it is the same. Using the first two terms and then the last two terms we get two expressions for the common ratio r.
r=\dfrac{G}{a}\quad\text{(i)},\qquad r=\dfrac{b}{G}\quad\text{(ii)}
Equate (i) and (ii) and cross-multiply.
\dfrac{G}{a}=\dfrac{b}{G}\;\Longrightarrow\;G^2=ab
Take the square root.
G=\pm\sqrt{ab}

ResultG=\pm\sqrt{ab}

The geometric mean of two numbers is the square root of their product.

Example 17

Find the geometric mean of each of the following pairs of numbers. (i) 9 and 4 (ii) -\dfrac32 and -\dfrac{27}8.

(i) Here a=9 and b=4. Both are positive, so we take the positive root.

Use G=\sqrt{ab}.
G=\sqrt{9\times4}=\sqrt{36}=6

(ii) Here a=-\dfrac32 and b=-\dfrac{27}8. Both are negative, so we take the negative root.

The product of two negative numbers is positive.
ab=\left(-\dfrac32\right)\left(-\dfrac{27}8\right)=\dfrac{81}{16}
Take the square root, with the sign chosen by the key facts.
G=-\sqrt{\dfrac{81}{16}}=-\dfrac94
Check that -\dfrac32,-\dfrac94,-\dfrac{27}8 is a G.P.: the ratio is \dfrac{-9/4}{-3/2}=\dfrac32 and \dfrac{-27/8}{-9/4}=\dfrac32.

\text{(i) } 6\qquad\text{(ii) } -\dfrac94

Example 18

Find two geometric means between 81 and 3.

The sequence is 81,\_,\_,3. It has four terms, so a_1=81 and a_4=3. Use a_n=a_1r^{n-1} to find r.
a_4=a_1r^3\;\Longrightarrow\;3=81\,r^3
Solve for r^3 and take the cube root.
r^3=\dfrac3{81}=\dfrac1{27}=\left(\dfrac13\right)^3\;\Longrightarrow\;r=\dfrac13
Multiply by r to find the missing means.
a_2=a_1r=81\left(\dfrac13\right)=27,\qquad a_3=a_1r^2=81\left(\dfrac13\right)^2=9

27,\ 9

Example 19

A vacuum pump removes \dfrac15 of the air from a sealed container on each stroke of its piston. What percent of the air remains after five strokes of the piston?

Let 1 represent the original amount of air. The first stroke removes \dfrac15, so 1-\dfrac15=\dfrac45 of the air remains.
The second stroke removes \dfrac15 of the air that is left, so the amount remaining after two strokes is \dfrac45\left(1-\dfrac15\right).
\dfrac45\cdot\dfrac45=\dfrac{16}{25}
The amounts remaining form a geometric sequence with common ratio \dfrac45. Write them in a table: a_1=1 is the amount before any stroke, a_2 after one stroke, and so on, so a_6 is the amount after five strokes.

Number of strokes	0	1	2	3	4	5
Amount of air remaining	1	\dfrac45	\dfrac{16}{25}	\dots	\dots	\dots
Term	a_1	a_2	a_3	a_4	a_5	a_6

Number of strokes	Amount of air remaining	Term
0	1	a_1
1	\dfrac45	a_2
2	\dfrac{16}{25}	a_3
3	\dots	a_4
4	\dots	a_5
5	\dots	a_6

Use a_n=a_1r^{n-1} with a_1=1, r=\dfrac45 and n=6.
a_6=1\cdot\left(\dfrac45\right)^5=\dfrac{4^5}{5^5}=\dfrac{1024}{3125}=0.32768
Convert the fraction to a percent.
0.32768\times100=32.768\%

\dfrac{1024}{3125}=0.32768\approx32.8\%\text{ of the air remains}

4.6, 4.6.1Geometric seriesOften examined

The sum of the terms of a geometric sequence is called a geometric series.

Sum of the first n terms of a geometric sequence

Derivation The sum of n terms of a G.P.

Aim. Prove S_n=\dfrac{a_1(1-r^n)}{1-r} for r\ne1.

Write the terms of the geometric sequence: a_1,\ a_1r,\ a_1r^2,\ a_1r^3,\dots,a_1r^{n-1}. Their sum is the geometric series S_n.
S_n=a_1+a_1r+a_1r^2+a_1r^3+\dots+a_1r^{n-2}+a_1r^{n-1}\qquad(1)
Multiply both sides of (1) by r. Every term moves up one power of r.
rS_n=a_1r+a_1r^2+a_1r^3+\dots+a_1r^{n-1}+a_1r^{n}\qquad(2)
Subtract (2) from (1). All the terms in the middle cancel, leaving only the first term of (1) and the last term of (2).
S_n-rS_n=a_1-a_1r^n
Factor both sides.
S_n(1-r)=a_1(1-r^n)
Divide both sides by 1-r.
S_n=\dfrac{a_1(1-r^n)}{1-r}

ResultS_n=\dfrac{a_1\left(1-r^{n}\right)}{1-r},\qquad r\neq1

When r=1 the denominator 1-r is zero, so the formula cannot be used.

Example 20

Find the sum of the first 7 terms of the geometric sequence 3,15,75,375,\dots

Read off the data. The common ratio is \dfrac{15}3=5.
a_1=3,\qquad n=7,\qquad r=5
Since |r|>1 use S_n=\dfrac{a_1(r^n-1)}{r-1}.
S_7=\dfrac{3\left(5^7-1\right)}{5-1}
Use 5^7=78125.
S_7=\dfrac{3(78125-1)}{4}=\dfrac{3\times78124}{4}=3\times19531=58593

58593

Another form of the formula

Another form of S_n can be used when the number of terms is not known but the last term a_n is. It follows from a_n=a_1r^{n-1}.

Multiply a_n=a_1r^{n-1} by r.
a_nr=a_1r^{n-1}\cdot r=a_1r^n
Substitute a_1r^n=a_nr into the numerator of S_n=\dfrac{a_1-a_1r^n}{1-r}.
S_n=\dfrac{a_1-a_nr}{1-r}

Example 21

Find the sum of a geometric series for which a_1=48, a_n=3 and r=-\dfrac12.

We do not know n, so use S_n=\dfrac{a_1-a_nr}{1-r}.
S_n=\dfrac{48-3\left(-\tfrac12\right)}{1-\left(-\tfrac12\right)}
Simplify the top and the bottom.
S_n=\dfrac{48+\tfrac32}{1+\tfrac12}=\dfrac{\tfrac{99}2}{\tfrac32}=33

Example 22

Find a_1 in a geometric series where S_7=3279 and r=3.

Use S_n=\dfrac{a_1(r^n-1)}{r-1} with n=7 to get S_7 in terms of a_1.
S_7=\dfrac{a_1\left(3^7-1\right)}{3-1}
Substitute S_7=3279 and 3^7=2187.
3279=\dfrac{a_1(2187-1)}{2}=\dfrac{2186\,a_1}{2}=1093\,a_1
Solve for a_1.
a_1=\dfrac{3279}{1093}=3

a_1=3

4.6.2Infinite geometric seriesOften examined

The first swing of a pendulum measures 25\,\mathrm{cm}. The lengths of the successive swings form the geometric sequence 25,20,16,12.8,\dots Suppose the pendulum continues to swing back and forth indefinitely. Then the sequence above becomes an infinite geometric sequence. The total distance the pendulum travels can be expressed as the infinite geometric series

25+20+16+12.8+\dots

In the series a_1=25 and r=\dfrac{20}{25}=0.8. So the series can be written

25+25(0.8)^1+25(0.8)^2+25(0.8)^3+25(0.8)^4+\dots

Look for a pattern in the values of (0.8)^n as n increases.

n	1	10	50
(0.8)^n	0.8	0.107374	0.0000143

n	(0.8)^n
1	0.8
10	0.107374
50	0.0000143

In an infinite geometric series with |r|<1, as n increases without bound, r^n approaches 0. Substituting r^n\approx0 into S_n=\dfrac{a_1(1-r^n)}{1-r} gives the formula for the sum of an infinite geometric series.

Derivation The sum of an infinite geometric series

Aim. Find S_\infty for -1<r<1.

Start with the sum of n terms.
S_n=\dfrac{a_1\left(1-r^{n}\right)}{1-r}
When |r|<1 and n\to\infty, the power r^n approaches 0. So 1-r^n approaches 1.
r^n\to0\;\Longrightarrow\;1-r^n\to1
The sum of the infinite series is the value that S_n approaches.
S_\infty=\dfrac{a_1}{1-r}

ResultS_\infty=\dfrac{a_1}{1-r},\qquad -1<r<1

The infinite series has a finite sum, even though it has infinitely many terms, because the terms shrink fast enough.

Partial sums of the pendulum series

The sums S_n of 25+20+16+\dots rise quickly at first and then level off below the dashed line S=125. As n increases r^n approaches 0 and S_n approaches S_\infty=125.

Example 23

Find the total distance travelled by the pendulum before coming to rest, if its successive swings form the geometric series 25+20+16+12.8+\dots

The series is infinite with a_1=25 and r=0.8. Since |r|<1 the sum exists.
Use S_\infty=\dfrac{a_1}{1-r}.
S=\dfrac{25}{1-0.8}=\dfrac{25}{0.2}=125

125\,\mathrm{cm}

Example 24

Find the sum of the infinite geometric series \dfrac43-\dfrac23+\dfrac13-\dfrac16+\dots

To find r divide any term by its preceding term.
r=\dfrac{-2/3}{4/3}=-\dfrac12
Since |r|=\dfrac12<1 the sum exists. Use S_\infty=\dfrac{a_1}{1-r} with a_1=\dfrac43.
S_\infty=\dfrac{4/3}{1-\left(-\tfrac12\right)}=\dfrac{4/3}{3/2}=\dfrac43\times\dfrac23=\dfrac89

\dfrac89

Example 25

Find the fractional notation for 0.63636363\dots

Express the decimal as a sum of terms.
0.63636363\dots=0.63+0.0063+0.000063+\dots
This is an infinite geometric series with a_1=0.63 and r=\dfrac{0.0063}{0.63}=0.01. Since |r|<1 it has a sum.
S_\infty=\dfrac{a_1}{1-r}=\dfrac{0.63}{1-0.01}=\dfrac{0.63}{0.99}=\dfrac{63}{99}
Cancel the common factor 9.
\dfrac{63}{99}=\dfrac7{11}

\dfrac{63}{99}=\dfrac7{11}

Worked example: total distance of a bouncing ball

A ball is dropped from a height of 20\,\mathrm{m}. After each fall it rebounds to \tfrac35 of the height it fell. Find the total distance it travels before coming to rest.

The ball falls 20\,\mathrm{m} once. Each rebound height is \tfrac35 of the previous one, and each rebound is travelled twice (up and down).

The first fall is a single distance. After that, each rebound is travelled up and then down, so every rebound height counts twice.
\text{total}=20+2\,(h_1+h_2+h_3+\dots)
The rebound heights form an infinite G.P.: h_1=\tfrac35(20)=12 and each next height is \tfrac35 of the one before.
12+7.2+4.32+\dots,\qquad a_1=12,\ r=\tfrac35
Since |r|<1 the sum exists. Use S_\infty=\dfrac{a_1}{1-r}.
S_\infty=\frac{12}{1-\tfrac35}=\frac{12}{\tfrac25}=30
Put this into the total.
\text{total}=20+2(30)=80\ \mathrm{m}
Common mistake: counting the rebounds once (giving 50\,\mathrm{m}) or counting the first fall twice.

80\,\mathrm{m}

4.7, 4.7.1Harmonic sequenceOften examined

Harmonic sequence (H.P.): A sequence of numbers is called a harmonic sequence or harmonic progression (H.P.) if the reciprocals of its terms are in arithmetic progression.

For example, the sequence 1,\dfrac14,\dfrac17,\dfrac1{10},\dots is a harmonic sequence because the reciprocals of its terms are 1,4,7,10,\dots, which form an arithmetic sequence (common difference 3).

The nth term of a harmonic sequence

Derivation The nth term of an H.P.

Aim. Find \dfrac1{a_n} for the harmonic sequence \dfrac1{a_1},\dfrac1{a_1+d},\dfrac1{a_1+2d},\dots

The sequence is an H.P. because the reciprocals of its terms, a_1,\ a_1+d,\ a_1+2d,\dots, are in A.P. with first term a_1 and common difference d.
The general term of that A.P. is
a_n=a_1+(n-1)d
The term of the H.P. is the reciprocal of this.
\dfrac1{a_n}=\dfrac1{a_1+(n-1)d}

Result\dfrac1{a_n}=\dfrac1{a_1+(n-1)d}\quad(\text{in H.P.})

Here a_1 and d are the first term and the common difference of the corresponding A.P. (the reciprocals), not of the H.P. itself.

Example 26

Find the 9th term of the H.P. \dfrac12,\dfrac17,\dfrac1{12},\dfrac1{17},\dots

Take reciprocals. The numbers 2,7,12,17,\dots are in A.P.
Read off the A.P. data: a_1=2, d=7-2=5, n=9.
Find the 9th term of the A.P.
a_9=a_1+(9-1)d=2+8(5)=42
The 9th term of the H.P. is the reciprocal.
\dfrac1{a_9}=\dfrac1{42}

\dfrac1{42}

Example 27

Find the harmonic sequence whose fourth term is \dfrac1{13} and eleventh term is \dfrac1{25}.

The reciprocals are in A.P., so the fourth and eleventh terms of the A.P. are 13 and 25.
a_4=13\;\Longrightarrow\;a_1+3d=13\quad\text{(i)}\qquad a_{11}=25\;\Longrightarrow\;a_1+10d=25\quad\text{(ii)}
Subtract (i) from (ii) to remove a_1.
7d=12\;\Longrightarrow\;d=\dfrac{12}7
Put d in (i) to find a_1.
a_1=13-3\left(\dfrac{12}7\right)=\dfrac{91-36}7=\dfrac{55}7
Write the first terms of the A.P.
\dfrac{55}7,\ \dfrac{67}7,\ \dfrac{79}7,\ 13,\ \dots
Take reciprocals to get the harmonic sequence.
\dfrac7{55},\ \dfrac7{67},\ \dfrac7{79},\ \dfrac1{13},\ \dots

\dfrac7{55},\ \dfrac7{67},\ \dfrac7{79},\ \dfrac1{13},\ \dots

4.7.2Harmonic mean

Harmonic mean (H.M.): A number H is the harmonic mean between two numbers a and b if a,\,H,\,b are in H.P.

Derivation The harmonic mean of two numbers

Aim. Find H when a,H,b are in H.P.

If a,H,b are in H.P. then their reciprocals \dfrac1a,\dfrac1H,\dfrac1b are in A.P.
In an A.P. the common difference is the same between neighbours.
\dfrac1H-\dfrac1a=\dfrac1b-\dfrac1H
Collect the \dfrac1H terms on one side.
\dfrac1H+\dfrac1H=\dfrac1a+\dfrac1b\;\Longrightarrow\;\dfrac2H=\dfrac{a+b}{ab}
Invert both sides and multiply by 2.
H=\dfrac{2ab}{a+b}

ResultH=\dfrac{2ab}{a+b}

This is the formula for the harmonic mean between a and b.

Example 28

Find the harmonic mean between 15 and 7.

Put a=15 and b=7 into H=\dfrac{2ab}{a+b}.
H.M.=\dfrac{2(15)(7)}{15+7}=\dfrac{210}{22}
Cancel the common factor 2.
\dfrac{210}{22}=\dfrac{105}{11}

\dfrac{105}{11}

4.7.3Relations between arithmetic, geometric and harmonic meansOften examined

Let A, G and H be the arithmetic, geometric and harmonic means between two numbers a and b. We know

Mean	Formula
Arithmetic mean	A=\dfrac{a+b}{2}
Geometric mean	G=\sqrt{ab}
Harmonic mean	H=\dfrac{2ab}{a+b}

(i) A > G > H

Derivation Show that A > G > H for two positive numbers

Aim. If a and b are two positive numbers, show that A>G>H.

First A>G. This is the statement \dfrac{a+b}2>\sqrt{ab}. Multiply by 2.
a+b>2\sqrt{ab}
Move everything to one side.
a+b-2\sqrt{ab}>0
The left side is a perfect square, because a=(\sqrt a)^2 and b=(\sqrt b)^2.
\left(\sqrt a-\sqrt b\right)^2>0
A square of a non-zero number is always positive, so this is always true. Every step can be read backwards, so the first statement is true too.
\therefore\;A>G
Next G>H. This is the statement \sqrt{ab}>\dfrac{2ab}{a+b}. Multiply by \dfrac{a+b}{\sqrt{ab}} (a positive number).
a+b>\dfrac{2ab}{\sqrt{ab}}=2\sqrt{ab}
This is the same inequality as before.
a+b-2\sqrt{ab}>0\;\Longrightarrow\;\left(\sqrt a-\sqrt b\right)^2>0,\ \text{always true}
So G>H. Put the two results together.
\therefore\;G>H

ResultA>G>H

The arithmetic mean is the largest and the harmonic mean the smallest of the three means of two positive numbers. (The squares are strictly positive when a\neq b; if a=b then all three means equal a.)

(ii) A × H = G²

Derivation Show that A × H = G²

Aim. Prove that the product of the arithmetic and harmonic means equals the square of the geometric mean.

Start with the left-hand side and substitute the formulas for A and H.
\text{L.H.S.}=A\times H=\dfrac{a+b}{2}\times\dfrac{2ab}{a+b}
Cancel 2 and a+b.
=ab
Write ab as the square of \sqrt{ab}.
=\left(\sqrt{ab}\right)^2=G^2=\text{R.H.S.}

ResultA\times H=G^2

G is the geometric mean of A and H.

Example 29

Find the arithmetic, geometric and harmonic means of 24 and 16. Also show that AH=G^2.

Here a=24 and b=16. Arithmetic mean:
A=\dfrac{a+b}{2}=\dfrac{24+16}{2}=20
Geometric mean (both numbers positive, so the positive root). Note 384=64\times6.
G=\sqrt{ab}=\sqrt{24\times16}=\sqrt{384}=8\sqrt6
Harmonic mean:
H=\dfrac{2ab}{a+b}=\dfrac{2(24)(16)}{24+16}=\dfrac{768}{40}=\dfrac{96}5
Left-hand side of AH=G^2:
A\times H=20\times\dfrac{96}{5}=384
Right-hand side:
G^2=\left(8\sqrt6\right)^2=64\times6=384
The two sides are equal. We also see A=20>G\approx19.6>H=19.2.
\therefore\;A\times H=G^2

A=20,\quad G=8\sqrt6,\quad H=\dfrac{96}{5}

4.8, 4.8.1Sigma notationOften examined

A sequence is simply an ordered list. For example, when a baseball coach writes a batting order, a sequence is being formed. When the members of a sequence are numbers we can find their sum, and such a sum is called a series.

Sigma notation

When the general term of a sequence is known, the Greek letter \Sigma (sigma) can be used to write a series. For example, the sum of the first four terms of the sequence 3,5,7,9,\dots,2k+1 can be written using sigma (summation) notation as

\sum_{k=1}^{4}(2k+1)

Read as: 'the sum as k goes from 1 to 4 of (2k+1)'.

The letter k is called the index of summation. The numbers below and above the sigma are the lowest and highest values of k. To expand the sum, put k=1,2,3,4 into the general term and add: 3+5+7+9.

Index of summation: The letter, here k, that runs through the values from the lower limit to the upper limit in a sigma sum.
Sigma notation: \sum_{k=m}^{n}f(k) means f(m)+f(m+1)+\dots+f(n).

Example 30

Find the following sums. (a) \displaystyle\sum_{k=1}^{5}k^2 (b) \displaystyle\sum_{k=1}^{4}(-1)^k(2k) (c) \displaystyle\sum_{k=0}^{3}\left(2^k+5\right)

(a) Evaluate k^2 for all integers from 1 to 5 and then add.

Expand.
\sum_{k=1}^{5}k^2=1^2+2^2+3^2+4^2+5^2=1+4+9+16+25=55

(b) Put k=1,2,3,4 into (-1)^k(2k). The factor (-1)^k is -1 for odd k and +1 for even k.

Expand.
\sum_{k=1}^{4}(-1)^k(2k)=(-1)^1(2\cdot1)+(-1)^2(2\cdot2)+(-1)^3(2\cdot3)+(-1)^4(2\cdot4)=-2+4-6+8=4

(c) This sum starts at k=0, so it has four terms: k=0,1,2,3.

Expand.
\sum_{k=0}^{3}\left(2^k+5\right)=(2^0+5)+(2^1+5)+(2^2+5)+(2^3+5)=6+7+9+13=35

\text{(a) } 55\qquad\text{(b) } 4\qquad\text{(c) } 35

Example 31

Write sigma notation for the sum. (a) 1+4+9+16+25 (b) -1+3-5+7 (c) 3+9+27+81+\dots

(a) This is a sum of squares, 1^2+2^2+3^2+4^2+5^2. So the general term is k^2, and there are five terms.

\sum_{k=1}^{5}k^2

(b) Except for the alternating signs, this is the sum of the first four positive odd numbers. The expression 2k-1 is a formula for the kth positive odd number, and (-1)^k=1 when k is even and -1 when k is odd. So the general term is (-1)^k(2k-1), beginning with k=1.

\sum_{k=1}^{4}(-1)^k(2k-1)

(c) This is the sum of powers of 3, and it is an infinite series. We use the symbol \infty for infinity.

\sum_{k=1}^{\infty}3^k

4.8.2Some important resultsOften examined

The sum of the first n natural numbers, the sum of the squares of the first n natural numbers and the sum of the cubes of the first n natural numbers are written in sigma notation as

\sum_{k=1}^{n}k=1+2+3+4+\dots+n

\sum_{k=1}^{n}k^2=1^2+2^2+3^2+4^2+\dots+n^2

\sum_{k=1}^{n}k^3=1^3+2^3+3^3+4^3+\dots+n^3

To find formulas for these we first evaluate \sum_{k=1}^{n}\left[k^m-(k-1)^m\right] for any positive integer m.

Derivation A telescoping sum

Aim. Show that \displaystyle\sum_{k=1}^{n}\left[k^m-(k-1)^m\right]=n^m.

Write out the terms for k=1,2,3,\dots,n.
\left(1^m-0^m\right)+\left(2^m-1^m\right)+\left(3^m-2^m\right)+\dots+\left[(n-1)^m-(n-2)^m\right]+\left[n^m-(n-1)^m\right]
Remove the brackets. Each term -k^m cancels with the next term +k^m.
=1^m-0^m+2^m-1^m+3^m-2^m+\dots+(n-1)^m-(n-2)^m+n^m-(n-1)^m
Only n^m is left (and 0^m=0); all the other terms cancel out.
=n^m

Result\sum_{k=1}^{n}\left[k^m-(k-1)^m\right]=n^m\qquad\text{(i)}

This is called a telescoping sum, because it collapses like a folding telescope. We use (i) with m=1,2,3 to get the three formulas.

Derivation Sum of n ones (m = 1)

Aim. Put m=1 in (i).

Substitute m=1.
\sum_{k=1}^{n}\left[k^1-(k-1)^1\right]=n^1
Simplify the bracket: k-(k-1)=1.
\sum_{k=1}^{n}[k-k+1]=n\;\Longrightarrow\;\sum_{k=1}^{n}1=n

Result\sum_{k=1}^{n}1=n

This means 1+1+1+\dots+1=n (n times).

Derivation Sum of the first n natural numbers (m = 2)

Aim. Prove \displaystyle\sum_{k=1}^{n}k=\dfrac{n(n+1)}2.

Put m=2 in (i).
\sum_{k=1}^{n}\left[k^2-(k-1)^2\right]=n^2
Expand (k-1)^2=k^2-2k+1 and simplify the bracket.
\sum_{k=1}^{n}\left[k^2-k^2+2k-1\right]=\sum_{k=1}^{n}(2k-1)=n^2
Split the sum into two sums and take out the constant 2.
2\sum_{k=1}^{n}k-\sum_{k=1}^{n}1=n^2
Use \sum_{k=1}^{n}1=n.
2\sum_{k=1}^{n}k-n=n^2\;\Longrightarrow\;2\sum_{k=1}^{n}k=n^2+n
Divide by 2 and factor.
\sum_{k=1}^{n}k=\dfrac{n^2+n}2=\dfrac{n(n+1)}2

Result\sum_{k=1}^{n}k=\dfrac{n(n+1)}2

Derivation Sum of squares of the first n natural numbers (m = 3)

Aim. Prove \displaystyle\sum_{k=1}^{n}k^2=\dfrac{n(n+1)(2n+1)}6.

Put m=3 in (i).
\sum_{k=1}^{n}\left[k^3-(k-1)^3\right]=n^3
Expand (k-1)^3=k^3-3k^2+3k-1 and simplify the bracket.
\sum_{k=1}^{n}\left[3k^2-3k+1\right]=n^3
Split into three sums.
3\sum_{k=1}^{n}k^2-3\sum_{k=1}^{n}k+\sum_{k=1}^{n}1=n^3
Use \sum k=\dfrac{n(n+1)}2 and \sum1=n.
3\sum_{k=1}^{n}k^2-\dfrac{3n(n+1)}2+n=n^3
Make 3\sum k^2 the subject and put everything over 2.
3\sum_{k=1}^{n}k^2=n^3-n+\dfrac{3n(n+1)}2=\dfrac{2n^3-2n+3n^2+3n}2=\dfrac{2n^3+3n^2+n}2
Factor: 2n^3+3n^2+n=n(2n^2+3n+1)=n(n+1)(2n+1).
3\sum_{k=1}^{n}k^2=\dfrac{n(n+1)(2n+1)}2
Divide by 3.
\sum_{k=1}^{n}k^2=\dfrac{n(n+1)(2n+1)}6

Result\sum_{k=1}^{n}k^2=\dfrac{n(n+1)(2n+1)}6

Similarly (taking m=4 in (i)) we can prove the sum of cubes:

\sum_{k=1}^{n}k^3=\left[\dfrac{n(n+1)}2\right]^2

The book states this result without working and asks for the proof in Exercise 4.7.

The four standard sums.
Sum	Result
\displaystyle\sum_{k=1}^{n}1	n
\displaystyle\sum_{k=1}^{n}k	\dfrac{n(n+1)}2
\displaystyle\sum_{k=1}^{n}k^2	\dfrac{n(n+1)(2n+1)}6
\displaystyle\sum_{k=1}^{n}k^3	\left[\dfrac{n(n+1)}2\right]^2

Example 32

Find the sum of the n terms of the series 1\cdot2+2\cdot3+3\cdot4+\dots

Find the general term. The first factors are 1,2,3,\dots with general term k. The second factors are 2,3,4,\dots with general term k+1. So the kth term is
T_k=k(k+1)=k^2+k
Take the summation of both sides and split it.
\sum_{k=1}^{n}T_k=\sum_{k=1}^{n}k^2+\sum_{k=1}^{n}k
Use the standard sums.
=\dfrac{n(n+1)(2n+1)}6+\dfrac{n(n+1)}2
Take out the common factor \dfrac{n(n+1)}2.
=\dfrac{n(n+1)}2\left[\dfrac{2n+1}3+1\right]
Add inside the bracket: \dfrac{2n+1}3+1=\dfrac{2n+4}3.
=\dfrac{n(n+1)}2\cdot\dfrac{2(n+2)}3=\dfrac{n(n+1)(n+2)}3

\sum_{k=1}^{n}T_k=\dfrac{n(n+1)(n+2)}3

Example 33

Find the sum to n terms of the series whose nth term is n^2+4n+1.

Replace n by k to get the general term.
T_k=k^2+4k+1
Take the summation and split it. Take out the constant 4.
\sum T_k=\sum k^2+4\sum k+\sum1
Use the standard sums.
=\dfrac{n(n+1)(2n+1)}6+4\cdot\dfrac{n(n+1)}2+n
Take out the common factor n and put everything over 6.
=\dfrac n6\left[(n+1)(2n+1)+12(n+1)+6\right]
Expand: (n+1)(2n+1)=2n^2+3n+1.
=\dfrac n6\left[2n^2+3n+1+12n+12+6\right]
Collect like terms.
=\dfrac{n\left(2n^2+15n+19\right)}6

\sum_{k=1}^{n}T_k=\dfrac{n\left(2n^2+15n+19\right)}6

Worked example: sigma notation with the standard results

Find \displaystyle\sum_{k=1}^{n}(2k^2-3k+1) as a formula in n and check it for n=3.

Split the sum and take out the constant factors.
\sum_{k=1}^{n}(2k^2-3k+1)=2\sum k^2-3\sum k+\sum 1
Use \sum k^2=\dfrac{n(n+1)(2n+1)}6, \sum k=\dfrac{n(n+1)}2 and \sum1=n.
=\frac{n(n+1)(2n+1)}3-\frac{3n(n+1)}2+n
Put everything over 6 and take out the common factor n.
=\frac{2n(n+1)(2n+1)-9n(n+1)+6n}6=\frac{n\,[\,4n^2+6n+2-9n-9+6\,]}6=\frac{n(4n^2-3n-1)}6
Factorise the quadratic: 4n^2-3n-1=(4n+1)(n-1).
=\frac{n(n-1)(4n+1)}6
Check for n=3. Direct sum: k=1,2,3 give 0,\,3,\,10, total 13. Formula:
\frac{3(2)(13)}6=13\ \checkmark

\dfrac{n(n-1)(4n+1)}6

4.9, 4.9.1Arithmetico-geometric seriesOften examined

In mathematics, an arithmetico-geometric sequence is the result of term-by-term multiplication of a geometric progression with the corresponding terms of an arithmetic progression. The nth term of an arithmetico-geometric sequence is the product of the nth term of an arithmetic sequence and the nth term of a geometric sequence. Such sequences arise in various applications, such as the computation of expected values in statistics, and in other fields.

For instance the sequence

\dfrac01,\ \dfrac12,\ \dfrac24,\ \dfrac38,\ \dfrac4{16},\ \dfrac5{32},\ \dots

is an arithmetico-geometric sequence. The arithmetic component (0,1,2,3,4,5,\dots) appears in the numerators and the geometric one (1,2,4,8,16,32,\dots) in the denominators. The summation of such an infinite sequence is known as an arithmetico-geometric series.

Arithmetico-geometric sequence: A sequence whose nth term is T_n=A_nG_n, where A_n is the nth term of an arithmetic sequence and G_n is the nth term of a geometric sequence.
Arithmetico-geometric series: The sum of the terms of an arithmetico-geometric sequence.

Terms of the sequence

The first few terms of an arithmetico-geometric sequence made from an arithmetic progression with initial value a and common difference d, and a geometric progression with initial value b and common ratio r, are:

\begin{aligned} T_1&=ab=A_1G_1\\ T_2&=(a+d)\,br=A_2G_2\\ T_3&=(a+2d)\,br^2=A_3G_3\\ &\;\;\vdots\\ T_n&=\underbrace{[a+(n-1)d]}_{A_n}\,\underbrace{b\,r^{\,n-1}}_{G_n}=A_nG_n\end{aligned}

Worked example in the book (nth term of 0/1, 1/2, 2/4, ...)

Find the nth term of \dfrac01,\dfrac12,\dfrac24,\dfrac38,\dfrac4{16},\dfrac5{32},\dots

The numerators 0,1,2,3,\dots are an A.P. with first term a=0 and d=1. The reciprocals of the denominators, 1,\dfrac12,\dfrac14,\dfrac18,\dots, are a G.P. with first term b=1 and r=\dfrac12.
a=0,\quad d=1,\quad b=1,\quad r=\dfrac12
Put these into T_n=[a+(n-1)d]\,br^{n-1}.
T_n=[0+(n-1)1]\cdot1\cdot\left(\dfrac12\right)^{n-1}
Simplify.
T_n=(n-1)\left(\dfrac12\right)^{n-1}=\dfrac{n-1}{2^{n-1}}
Check: n=4 gives \dfrac{3}{2^3}=\dfrac38, the fourth term.

T_n=\dfrac{n-1}{2^{\,n-1}}

4.9.2Sum of the n termsOften examined

The sum of the first n terms of an arithmetico-geometric sequence has the form

S_n=\sum_{k=1}^{n}T_k=\sum_{k=1}^{n}[a+(k-1)d]\,br^{k-1}=ab+(a+d)br+(a+2d)br^2+\dots+[a+(n-1)d]\,br^{n-1}\qquad(\mathrm i)

that is, S_n=A_1G_1+A_2G_2+A_3G_3+\dots+A_nG_n. This sum can be written in closed form. The book proves it with b=1, as it explains in the note below.

Derivation Sum of n terms of an arithmetico-geometric series

Aim. Prove S_n=\dfrac a{1-r}+\dfrac{dr\,(1-r^n)}{(1-r)^2}-\dfrac{(a+nd)\,r^n}{1-r}.

Put b=1 in (i).
S_n=a+(a+d)r+(a+2d)r^2+\dots+[a+(n-1)d]\,r^{n-1}\qquad(\mathrm{ii})
Multiply both sides of (ii) by r. Every term moves up one power of r.
rS_n=ar+(a+d)r^2+(a+2d)r^3+\dots+[a+(n-2)d]\,r^{n-1}+[a+(n-1)d]\,r^n\qquad(\mathrm{iii})
Subtract (iii) from (ii). Look at the terms column by column: the terms in r^k (for k=1,\dots,n-1) differ by d, since (a+kd)-(a+(k-1)d)=d. The first term a and the last term of (iii) do not pair up.
S_n-rS_n=a+d\left(r+r^2+r^3+\dots+r^{n-1}\right)-[a+(n-1)d]\,r^n
Write -[a+(n-1)d]r^n=-(a+nd)r^n+dr^n. The extra dr^n completes the bracket to r+r^2+\dots+r^{n-1}+r^n.
(1-r)S_n=a+d\left(r+r^2+r^3+\dots+r^{n-1}+r^n\right)-(a+nd)\,r^n
The bracket is a G.P. with first term r and ratio r, with n terms. Its sum is \dfrac{r(1-r^n)}{1-r}.
(1-r)S_n=a+dr\cdot\dfrac{1-r^n}{1-r}-(a+nd)\,r^n
Divide by 1-r (with r\ne1).
S_n=\dfrac a{1-r}+dr\,\dfrac{1-r^n}{(1-r)^2}-\dfrac{(a+nd)\,r^n}{1-r}\qquad(\mathrm{iv})

ResultS_n=\dfrac a{1-r}+\dfrac{dr\,(1-r^n)}{(1-r)^2}-\dfrac{(a+nd)\,r^n}{1-r}

Here a is the first term and d the common difference of the arithmetic series, and r is the common ratio of the geometric series. The three parts are \dfrac a{1-r}, the part from the common difference, and the end correction \dfrac{(a+nd)r^n}{1-r}.

4.9.3Sum to infinite terms of arithmetico-geometric seriesOften examined

Let |r|<1. We know that r^n\to0 as n\to\infty.

Derivation The sum to infinity of an arithmetico-geometric series

Aim. Find S_\infty for |r|<1.

Start with the formula for the sum of n terms.
S_n=\dfrac a{1-r}+\dfrac{dr\,(1-r^n)}{(1-r)^2}-\dfrac{(a+nd)\,r^n}{1-r}
As n\to\infty we have r^n\to0, so 1-r^n\to1. The last term also goes to 0, because nr^n\to0 when |r|<1 (the power of r shrinks faster than n grows).
S_\infty=\dfrac a{1-r}+\dfrac{dr}{(1-r)^2}

ResultS_\infty=\dfrac a{1-r}+\dfrac{dr}{(1-r)^2},\qquad|r|<1

This is the sum to infinity of an arithmetico-geometric series.

Example 34

Find the sum of 1+\dfrac32+\dfrac54+\dfrac78+\dots to n terms.

We use the formula for the sum of n terms of an arithmetico-geometric series.
S_n=\dfrac a{1-r}+dr\,\dfrac{(1-r^n)}{(1-r)^2}-\dfrac{(a+nd)\,r^n}{1-r}\qquad(\mathrm a)
Rearrange the series so each term is an A.P. term times a G.P. term.
1\cdot\dfrac11+3\cdot\dfrac12+5\cdot\dfrac14+7\cdot\dfrac18+\dots
Guess from the pattern: 1,3,5,7,\dots is an A.P. with a=1 and d=2, and \dfrac11,\dfrac12,\dfrac14,\dfrac18,\dots is a G.P. with r=\dfrac12.
a=1,\quad d=2,\quad r=\dfrac12
Substitute into (a).
S_n=\dfrac1{1-\tfrac12}+2\cdot\dfrac12\cdot\dfrac{1-\tfrac1{2^n}}{\left(1-\tfrac12\right)^2}-\dfrac{(1+2n)\tfrac1{2^n}}{1-\tfrac12}
Simplify each part: 1-\tfrac12=\tfrac12, \left(\tfrac12\right)^2=\tfrac14.
S_n=2+1\cdot\dfrac{1-\tfrac1{2^n}}{\tfrac14}-2(1+2n)\dfrac1{2^n}=2+4\left(1-\dfrac1{2^n}\right)-\dfrac{2(1+2n)}{2^n}
Expand and collect the terms with \dfrac1{2^n}.
S_n=2+4-\dfrac4{2^n}-\dfrac{2+4n}{2^n}=6-\dfrac1{2^n}\left(4+2+4n\right)=6-\dfrac2{2^n}(3+2n)
Write \dfrac2{2^n}=\dfrac1{2^{n-1}}.
S_n=6-\dfrac{2n+3}{2^{\,n-1}}
Check: n=1 gives 6-\dfrac{5}{1}=1, the first term; n=2 gives 6-\dfrac72=\dfrac52=1+\dfrac32.

S_n=6-\dfrac{2n+3}{2^{\,n-1}}

Example 35

Find the sum to infinity of the arithmetico-geometric series 1+\dfrac23+\dfrac39+\dfrac4{27}+\dfrac5{81}+\dots

Write the series so the A.P. and G.P. parts show.
1\times1+2\times\dfrac13+3\times\dfrac19+4\times\dfrac1{27}+5\times\dfrac1{81}+\dots
The numbers 1,2,3,4,5,\dots are in A.P. with a=1, d=1. The numbers 1,\dfrac13,\dfrac19,\dfrac1{27},\dfrac1{81},\dots are in G.P. with first term 1 and r=\dfrac{1/3}{1}=\dfrac13.
a=1,\quad d=1,\quad r=\dfrac13
Since |r|<1 the sum to infinity exists. Use S_\infty=\dfrac a{1-r}+\dfrac{dr}{(1-r)^2}.
S_\infty=\dfrac1{1-\tfrac13}+\dfrac{1\cdot\tfrac13}{\left(1-\tfrac13\right)^2}
Simplify: 1-\tfrac13=\tfrac23 and \left(\tfrac23\right)^2=\tfrac49.
S_\infty=\dfrac1{2/3}+\dfrac13\cdot\dfrac9{4}=\dfrac32+\dfrac34
Add.
S_\infty=\dfrac{6+3}{4}=\dfrac94

\dfrac94

4.10Applications of sequence and series

Sequences and series have their own importance in many areas of mathematics, such as finance, statistics, population growth and physics. Most of society and the reality around us is based upon sequence after sequence, changing and repeating themselves over and over again. Common examples of this are time and the calendrical system. Time (seconds, minutes, hours) always follows the same sequence, which always contains the same number of elements. Our lives are ruled by sequences such as the routines that we follow every day, without knowing their great importance in the structure and function of the modern world.

Example 36

Khalid is saving for a new car. He deposits Rs. 100{,}000 into his account and then each month he deposits Rs. 10{,}000 more than the month before. If the price of the car is Rs. 1{,}260{,}000, find (i) the amount Khalid has saved in four months, (ii) the time in which Khalid reaches his goal of Rs. 1{,}260{,}000.

(i) Khalid increases his deposit by the same amount every month, so we use an arithmetic series.

Read off the data. We want S_4.
a_1=100000,\quad d=10000,\quad n=4
Use S_n=\dfrac n2\left[2a_1+(n-1)d\right].
S_4=\dfrac42\left[2(100000)+(4-1)(10000)\right]
Simplify.
S_4=2\,[200000+30000]=2\times230000=460000

(ii) Now S_n=1{,}260{,}000 and we want n.

Put S_n=1260000, a_1=100000, d=10000 into the sum formula.
1260000=\dfrac n2\left[2(100000)+(n-1)10000\right]
Simplify inside the bracket: 200000+10000n-10000=190000+10000n.
1260000=\dfrac n2\left[190000+10000n\right]
Multiply by 2.
2520000=190000\,n+10000\,n^2
Divide by 10000 and bring everything to one side.
n^2+19n-252=0
Factor: look for two numbers with product -252 and sum 19, namely 28 and -9.
n^2+28n-9n-252=0\;\Longrightarrow\;n(n+28)-9(n+28)=0\;\Longrightarrow\;(n-9)(n+28)=0
So n=9 or n=-28. The number of months cannot be negative, so n=9.
n=9

\text{(i) Rs. } 460{,}000\qquad\text{(ii) the }9\text{th month}

Example 37

A new virus is on a remote area. On day one there were 10 people infected, with the number of new infections increasing at a rate of 40\% per day. (i) Find the expected number of infected people on the 7th day. (ii) Find the expected number of infected people during the week (7 days).

(i) The infection increases by a percentage each day, so this is a geometric sequence.

An increase of 40\% means each day has 100\%+40\%=140\% of the day before, so r=1.4.
a_1=10,\quad r=1.4,\quad a_7=\;?
Use a_n=a_1r^{n-1} with n=7.
a_7=a_1r^6=10(1.4)^6
Evaluate (1.4)^6=7.529536.
a_7=75.29536\approx75

(ii) The total number of infections in the week is S_7. Here r>1, so we use S_n=\dfrac{a_1(r^n-1)}{r-1}.

Substitute a_1=10, r=1.4, n=7.
S_7=\dfrac{10\left(1.4^7-1\right)}{1.4-1}
Evaluate (1.4)^7=10.5413504.
S_7=\dfrac{10(10.5413504-1)}{0.4}=\dfrac{95.413504}{0.4}=238.53376
Round to a whole number of people.
S_7\approx239

\text{(i) about }75\text{ people on the 7th day}\qquad\text{(ii) about }239\text{ people in the week}

Formula sheet

nth term of an A.P.

a_n=a_1+(n-1)d

a_1: first term; d: common difference; n: term number

Common difference

d=a_{n+1}-a_n

any term minus the term before it

Arithmetic mean

A=\dfrac{a+b}{2}

A: the A.M. between a and b (a,A,b in A.P.)

Sum of n terms of an A.P. (first and last term)

S_n=\dfrac n2\left[a_1+a_n\right]

a_n: last term

Sum of n terms of an A.P. (common difference)

S_n=\dfrac n2\left[2a_1+(n-1)d\right]

a_1: first term; d: common difference

nth term of a G.P.

a_n=a_1r^{\,n-1}

r: common ratio \dfrac{a_{n+1}}{a_n}

Geometric mean

G=\pm\sqrt{ab}

a,G,b in G.P.; positive root for positive a,b, negative root for negative a,b

Sum of n terms of a G.P.

S_n=\dfrac{a_1(1-r^n)}{1-r}=\dfrac{a_1(r^n-1)}{r-1},\qquad r\neq1

use the second form when |r|>1

Sum of a G.P. from the last term

S_n=\dfrac{a_1-a_nr}{1-r}

used when n is not known but the last term a_n is

Sum of an infinite G.P.

S_\infty=\dfrac{a_1}{1-r},\qquad -1<r<1

no sum exists when |r|>1

nth term of an H.P.

\dfrac1{a_n}=\dfrac1{a_1+(n-1)d}

a_1,d: first term and common difference of the A.P. of reciprocals

Harmonic mean

H=\dfrac{2ab}{a+b}

a,H,b in H.P.

Relations between the means

A>G>H,\qquad A\times H=G^2

a,b positive; for negative a,b: A<G<H

Telescoping sum

\sum_{k=1}^{n}\left[k^m-(k-1)^m\right]=n^m

m: any positive integer

Sum of n ones

\sum_{k=1}^{n}1=n

Sum of the first n natural numbers

\sum_{k=1}^{n}k=\dfrac{n(n+1)}2

Sum of squares of the first n natural numbers

\sum_{k=1}^{n}k^2=\dfrac{n(n+1)(2n+1)}6

Sum of cubes of the first n natural numbers

\sum_{k=1}^{n}k^3=\left[\dfrac{n(n+1)}2\right]^2

Properties of sigma

\sum(a+b)=\sum a+\sum b,\qquad \sum ca=c\sum a

c: a constant

nth term of an arithmetico-geometric sequence

T_n=[a+(n-1)d]\,b\,r^{\,n-1}=A_nG_n

a,d: A.P. first term and difference; b,r: G.P. first term and ratio

Sum of n terms of an arithmetico-geometric series (b = 1)

S_n=\dfrac a{1-r}+\dfrac{dr\,(1-r^n)}{(1-r)^2}-\dfrac{(a+nd)\,r^n}{1-r}

r\neq1

Sum to infinity of an arithmetico-geometric series

S_\infty=\dfrac a{1-r}+\dfrac{dr}{(1-r)^2},\qquad|r|<1

Key terms

Sequence: An ordered list of numbers; a function whose domain is the set of positive integers.
Finite and infinite sequence: A finite sequence has a last term; an infinite sequence continues without bound.
General term: The formula a_n for the nth term of a sequence.
Arithmetic sequence (A.P.): A sequence in which each term after the first is found by adding the same number to the preceding term.
Common difference: The number d added each time in an A.P.; any term minus the term before it.
Arithmetic mean: The number A with a,A,b in A.P.; A=\dfrac{a+b}2.
Series: The indicated sum of the terms of a sequence.
Arithmetic series: The sum of the terms of an arithmetic sequence; S_n=\dfrac n2[a_1+a_n].
Geometric sequence (G.P.): A sequence in which each term after the first is found by multiplying the previous term by the same number.
Common ratio: The constant multiplier r in a G.P.; any term divided by the term before it.
Geometric mean: The number G with a,G,b in G.P.; G=\pm\sqrt{ab}.
Geometric series: The sum of the terms of a geometric sequence.
Infinite geometric series: A geometric series with infinitely many terms; it has the sum \dfrac{a_1}{1-r} when |r|<1 and no sum when |r|>1.
Harmonic sequence (H.P.): A sequence whose reciprocals are in arithmetic progression.
Harmonic mean: The number H with a,H,b in H.P.; H=\dfrac{2ab}{a+b}.
Sigma notation: The notation \sum_{k=m}^{n}f(k) for the sum f(m)+f(m+1)+\dots+f(n).
Index of summation: The letter, usually k, that takes each whole value from the lower to the upper limit in a sigma sum.
Arithmetico-geometric sequence: A sequence whose nth term is the product of the nth term of an A.P. and the nth term of a G.P.
Arithmetico-geometric series: The sum of the terms of an arithmetico-geometric sequence.

Exercise 4.1p. 119

Finding terms of a sequence from its general term a_n (substitute n=1,2,3,\dots) and predicting a_n from the pattern of the first few terms.

In each of the following, the nth term of the sequence is given. In each case find the first 4 terms, the 10th term, a_{10} and the 15th term, a_{15}.

Solution

(i) a_n=3n+1

Substitute n=1,2,3,4 one by one in a_n.
\begin{aligned} a_{1}&=3(1)+1=4 \\ a_{2}&=3(2)+1=7 \\ a_{3}&=3(3)+1=10 \\ a_{4}&=3(4)+1=13 \end{aligned}
Substitute n=10 and n=15 in the same way.
\begin{aligned} a_{10}&=3(10)+1=31 \\ a_{15}&=3(15)+1=46 \end{aligned}

a_1=4,\ a_2=7,\ a_3=10,\ a_4=13,\ a_{10}=31,\ a_{15}=46

(ii) a_n=3n-1

Substitute n=1,2,3,4 one by one in a_n.
\begin{aligned} a_{1}&=3(1)-1=2 \\ a_{2}&=3(2)-1=5 \\ a_{3}&=3(3)-1=8 \\ a_{4}&=3(4)-1=11 \end{aligned}
Substitute n=10 and n=15 in the same way.
\begin{aligned} a_{10}&=3(10)-1=29 \\ a_{15}&=3(15)-1=44 \end{aligned}

a_1=2,\ a_2=5,\ a_3=8,\ a_4=11,\ a_{10}=29,\ a_{15}=44

(iii) a_n=\dfrac{n}{n+1}

Substitute n=1,2,3,4 one by one in a_n.
\begin{aligned} a_{1}&=\dfrac{1}{1+1}=\dfrac{1}{2} \\ a_{2}&=\dfrac{2}{2+1}=\dfrac{2}{3} \\ a_{3}&=\dfrac{3}{3+1}=\dfrac{3}{4} \\ a_{4}&=\dfrac{4}{4+1}=\dfrac{4}{5} \end{aligned}
Substitute n=10 and n=15 in the same way.
\begin{aligned} a_{10}&=\dfrac{10}{10+1}=\dfrac{10}{11} \\ a_{15}&=\dfrac{15}{15+1}=\dfrac{15}{16} \end{aligned}

a_1=\dfrac{1}{2},\ a_2=\dfrac{2}{3},\ a_3=\dfrac{3}{4},\ a_4=\dfrac{4}{5},\ a_{10}=\dfrac{10}{11},\ a_{15}=\dfrac{15}{16}

(iv) a_n=n^2+1

Substitute n=1,2,3,4 one by one in a_n.
\begin{aligned} a_{1}&=(1)^2+1=2 \\ a_{2}&=(2)^2+1=5 \\ a_{3}&=(3)^2+1=10 \\ a_{4}&=(4)^2+1=17 \end{aligned}
Substitute n=10 and n=15 in the same way.
\begin{aligned} a_{10}&=(10)^2+1=101 \\ a_{15}&=(15)^2+1=226 \end{aligned}

a_1=2,\ a_2=5,\ a_3=10,\ a_4=17,\ a_{10}=101,\ a_{15}=226

(v) a_n=n^2-2n

Substitute n=1,2,3,4 one by one in a_n.
\begin{aligned} a_{1}&=(1)^2-2(1)=-1 \\ a_{2}&=(2)^2-2(2)=0 \\ a_{3}&=(3)^2-2(3)=3 \\ a_{4}&=(4)^2-2(4)=8 \end{aligned}
Substitute n=10 and n=15 in the same way.
\begin{aligned} a_{10}&=(10)^2-2(10)=80 \\ a_{15}&=(15)^2-2(15)=195 \end{aligned}

a_1=-1,\ a_2=0,\ a_3=3,\ a_4=8,\ a_{10}=80,\ a_{15}=195

(vi) a_n=\dfrac{n^2-1}{n^2+1}

Substitute n=1,2,3,4 one by one in a_n.
\begin{aligned} a_{1}&=\dfrac{(1)^2-1}{(1)^2+1}=0 \\ a_{2}&=\dfrac{(2)^2-1}{(2)^2+1}=\dfrac{3}{5} \\ a_{3}&=\dfrac{(3)^2-1}{(3)^2+1}=\dfrac{4}{5} \\ a_{4}&=\dfrac{(4)^2-1}{(4)^2+1}=\dfrac{15}{17} \end{aligned}
Substitute n=10 and n=15 in the same way.
\begin{aligned} a_{10}&=\dfrac{(10)^2-1}{(10)^2+1}=\dfrac{99}{101} \\ a_{15}&=\dfrac{(15)^2-1}{(15)^2+1}=\dfrac{112}{113} \end{aligned}

a_1=0,\ a_2=\dfrac{3}{5},\ a_3=\dfrac{4}{5},\ a_4=\dfrac{15}{17},\ a_{10}=\dfrac{99}{101},\ a_{15}=\dfrac{112}{113}

(vii) a_n=\left(\dfrac{-1}{2}\right)^{n-1}

Substitute n=1,2,3,4 one by one in a_n.
\begin{aligned} a_{1}&=\left(\dfrac{-1}{2}\right)^{1-1}=1 \\ a_{2}&=\left(\dfrac{-1}{2}\right)^{2-1}=-\dfrac{1}{2} \\ a_{3}&=\left(\dfrac{-1}{2}\right)^{3-1}=\dfrac{1}{4} \\ a_{4}&=\left(\dfrac{-1}{2}\right)^{4-1}=-\dfrac{1}{8} \end{aligned}
Here \left(\dfrac{-1}{2}\right)^{n-1} alternates in sign: an even power of a negative number is positive and an odd power is negative.
Substitute n=10 and n=15 in the same way.
\begin{aligned} a_{10}&=\left(\dfrac{-1}{2}\right)^{10-1}=-\dfrac{1}{512} \\ a_{15}&=\left(\dfrac{-1}{2}\right)^{15-1}=\dfrac{1}{16384} \end{aligned}

a_1=1,\ a_2=-\dfrac{1}{2},\ a_3=\dfrac{1}{4},\ a_4=-\dfrac{1}{8},\ a_{10}=-\dfrac{1}{512},\ a_{15}=\dfrac{1}{16384}

(viii) a_n=(-1)^n n^2

Substitute n=1,2,3,4 one by one in a_n.
\begin{aligned} a_{1}&=(-1)^{1}(1)^2=-1 \\ a_{2}&=(-1)^{2}(2)^2=4 \\ a_{3}&=(-1)^{3}(3)^2=-9 \\ a_{4}&=(-1)^{4}(4)^2=16 \end{aligned}
The factor of -1 raised to n (or n+1) only decides the sign: it is -1 for odd n and +1 for even n in (-1)^n, and the opposite in (-1)^{n+1}.
Substitute n=10 and n=15 in the same way.
\begin{aligned} a_{10}&=(-1)^{10}(10)^2=100 \\ a_{15}&=(-1)^{15}(15)^2=-225 \end{aligned}

a_1=-1,\ a_2=4,\ a_3=-9,\ a_4=16,\ a_{10}=100,\ a_{15}=-225

(ix) a_n=(-1)^n(n+3)

Substitute n=1,2,3,4 one by one in a_n.
\begin{aligned} a_{1}&=(-1)^{1}(1+3)=-4 \\ a_{2}&=(-1)^{2}(2+3)=5 \\ a_{3}&=(-1)^{3}(3+3)=-6 \\ a_{4}&=(-1)^{4}(4+3)=7 \end{aligned}
The factor of -1 raised to n (or n+1) only decides the sign: it is -1 for odd n and +1 for even n in (-1)^n, and the opposite in (-1)^{n+1}.
Substitute n=10 and n=15 in the same way.
\begin{aligned} a_{10}&=(-1)^{10}(10+3)=13 \\ a_{15}&=(-1)^{15}(15+3)=-18 \end{aligned}

a_1=-4,\ a_2=5,\ a_3=-6,\ a_4=7,\ a_{10}=13,\ a_{15}=-18

(x) a_n=(-1)^{n+1}(3n-5)

Substitute n=1,2,3,4 one by one in a_n.
\begin{aligned} a_{1}&=(-1)^{1+1}(3(1)-5)=-2 \\ a_{2}&=(-1)^{2+1}(3(2)-5)=-1 \\ a_{3}&=(-1)^{3+1}(3(3)-5)=4 \\ a_{4}&=(-1)^{4+1}(3(4)-5)=-7 \end{aligned}
The factor of -1 raised to n (or n+1) only decides the sign: it is -1 for odd n and +1 for even n in (-1)^n, and the opposite in (-1)^{n+1}.
Substitute n=10 and n=15 in the same way.
\begin{aligned} a_{10}&=(-1)^{10+1}(3(10)-5)=-25 \\ a_{15}&=(-1)^{15+1}(3(15)-5)=40 \end{aligned}

a_1=-2,\ a_2=-1,\ a_3=4,\ a_4=-7,\ a_{10}=-25,\ a_{15}=40

Find the indicated term of the sequence

Solution

(i) a_n=4n-3;\ a_8

Put n=8 in the formula.
a_8=4(8)-3=32-3=29

a_8=29

(ii) a_n=5n+11;\ a_9

Put n=9.
a_9=5(9)+11=45+11=56

a_9=56

(iii) a_n=(3n+4)(2n-5);\ a_7

Put n=7 in each bracket.
a_7=(3\cdot7+4)(2\cdot7-5)=(25)(9)
Multiply.
a_7=225

a_7=225

(iv) a_n=(-1)^{n-1}(3.4n-17.3);\ a_{12}

Put n=12. The power n-1=11 is odd, so (-1)^{11}=-1.
a_{12}=(-1)^{11}(3.4\cdot12-17.3)
Evaluate the bracket: 3.4\times12=40.8 and 40.8-17.3=23.5.
a_{12}=-(23.5)=-23.5

a_{12}=-23.5

(v) a_n=4n^2(11n+31);\ a_{22}

Put n=22. First n^2=484, so 4n^2=1936.
4(22)^2=4(484)=1936
The bracket is 11(22)+31=242+31=273.
a_{22}=1936\times273
Multiply: 1936\times273=1936\times200+1936\times73=387200+141328.
a_{22}=528528

a_{22}=528528

(vi) a_n=\left(1+\dfrac1n\right)^2;\ a_{20}

Put n=20.
a_{20}=\left(1+\dfrac1{20}\right)^2=\left(\dfrac{21}{20}\right)^2
Square the numerator and the denominator.
a_{20}=\dfrac{441}{400}

a_{20}=\dfrac{441}{400}

(vii) a_n=\log10^{n};\ a_{43}

Use \log a^m=m\log a and \log10=1 (base 10), so \log10^n=n.
a_n=\log10^{n}=n\log10=n
Hence for n=43:
a_{43}=43

a_{43}=43

(viii) a_n=\ln e^{n};\ a_{67}

Use \ln e^n=n\ln e and \ln e=1, so \ln e^n=n.
a_n=\ln e^{n}=n\ln e=n
Hence for n=67:
a_{67}=67

a_{67}=67

Predict the general term or nth term, a_n, of the sequence.

Solution

(i) 1,3,5,7,9,\dots

These are the positive odd numbers. Each term is 2 more than the one before, so the formula starts with 2n. For n=1 the value 2n=2 is one too big (the term is 1), so subtract 1.
a_1=2(1)-1=1,\quad a_2=2(2)-1=3,\quad a_3=2(3)-1=5
The formula agrees with the given terms, e.g. a_5=2(5)-1=9.
a_n=2n-1

a_n=2n-1

(ii) 3,9,27,81,243,\dots

Each term is 3 times the term before it, so the terms are powers of 3.
3=3^1,\quad 9=3^2,\quad 27=3^3,\quad 81=3^4,\quad 243=3^5
The power equals the position n.
a_n=3^n

a_n=3^n

(iii) \sqrt2,\sqrt4,\sqrt6,\sqrt8,\sqrt{10},\dots

The numbers under the root sign are the even numbers 2,4,6,8,10, that is 2n for n=1,2,3,4,5.
a_1=\sqrt{2(1)},\quad a_2=\sqrt{2(2)},\quad a_3=\sqrt{2(3)}
So the general term is the square root of 2n.
a_n=\sqrt{2n}

a_n=\sqrt{2n}

(iv) 1.2,\ 2.3,\ 3.4,\ 4.5,\dots

Read the terms as decimals. The whole-number part of the nth term is n, and the decimal part is 0.1 times the next number n+1.
1.2=1+0.2,\quad 2.3=2+0.3,\quad 3.4=3+0.4,\quad 4.5=4+0.5
So the nth term is n plus 0.1(n+1). Combine the terms in n.
a_n=n+\dfrac{n+1}{10}=\dfrac{10n+n+1}{10}=\dfrac{11n+1}{10}
Check: a_1=\dfrac{12}{10}=1.2, a_2=\dfrac{23}{10}=2.3 and a_4=\dfrac{45}{10}=4.5.

a_n=\dfrac{11n+1}{10}

Exercise 4.2p. 122

Arithmetic sequences: a_n=a_1+(n-1)d with common difference d, and arithmetic means. For two unknowns a_1 and d, write the two given terms with the formula and solve the pair of equations.

Find the first four terms of each arithmetic sequence.

Solution

(i) a_1=4,\ d=3

The first term is given. Each next term of an arithmetic sequence is obtained by adding the common difference d to the term before it.
\begin{aligned} a_1&=4 \\ a_2&=a_1+d=4+3=7 \\ a_3&=a_2+d=7+3=10 \\ a_4&=a_3+d=10+3=13 \end{aligned}

a_1=4,\ a_2=7,\ a_3=10,\ a_4=13

(ii) a_1=7,\ d=5

The first term is given. Each next term of an arithmetic sequence is obtained by adding the common difference d to the term before it.
\begin{aligned} a_1&=7 \\ a_2&=a_1+d=7+5=12 \\ a_3&=a_2+d=12+5=17 \\ a_4&=a_3+d=17+5=22 \end{aligned}

a_1=7,\ a_2=12,\ a_3=17,\ a_4=22

(iii) a_1=16,\ d=-2

The first term is given. Each next term of an arithmetic sequence is obtained by adding the common difference d to the term before it.
\begin{aligned} a_1&=16 \\ a_2&=a_1+d=16+(-2)=14 \\ a_3&=a_2+d=14+(-2)=12 \\ a_4&=a_3+d=12+(-2)=10 \end{aligned}

a_1=16,\ a_2=14,\ a_3=12,\ a_4=10

(iv) a_1=38,\ d=-4

The first term is given. Each next term of an arithmetic sequence is obtained by adding the common difference d to the term before it.
\begin{aligned} a_1&=38 \\ a_2&=a_1+d=38+(-4)=34 \\ a_3&=a_2+d=34+(-4)=30 \\ a_4&=a_3+d=30+(-4)=26 \end{aligned}

a_1=38,\ a_2=34,\ a_3=30,\ a_4=26

(v) a_1=\dfrac{3}{4},\ d=\dfrac{1}{4}

The first term is given. Each next term of an arithmetic sequence is obtained by adding the common difference d to the term before it.
\begin{aligned} a_1&=\dfrac{3}{4} \\ a_2&=a_1+d=\dfrac{3}{4}+\dfrac{1}{4}=1 \\ a_3&=a_2+d=1+\dfrac{1}{4}=\dfrac{5}{4} \\ a_4&=a_3+d=\dfrac{5}{4}+\dfrac{1}{4}=\dfrac{3}{2} \end{aligned}

a_1=\dfrac{3}{4},\ a_2=1,\ a_3=\dfrac{5}{4},\ a_4=\dfrac{3}{2}

(vi) a_1=\dfrac{3}{8},\ d=\dfrac{5}{8}

The first term is given. Each next term of an arithmetic sequence is obtained by adding the common difference d to the term before it.
\begin{aligned} a_1&=\dfrac{3}{8} \\ a_2&=a_1+d=\dfrac{3}{8}+\dfrac{5}{8}=1 \\ a_3&=a_2+d=1+\dfrac{5}{8}=\dfrac{13}{8} \\ a_4&=a_3+d=\dfrac{13}{8}+\dfrac{5}{8}=\dfrac{9}{4} \end{aligned}

a_1=\dfrac{3}{8},\ a_2=1,\ a_3=\dfrac{13}{8},\ a_4=\dfrac{9}{4}

Find the next three terms of each arithmetic sequence.

Solution

(i) 5,9,13,\dots

Find d by subtracting a term from the one after it.
d=9-5=4\quad(\text{check: }13-9=4)
Add d=4 repeatedly to the last given term 13.
a_4=13+4=17,\quad a_5=17+4=21,\quad a_6=21+4=25

17,\ 21,\ 25

(ii) 11,14,17,\dots

Common difference:
d=14-11=3\quad(\text{check: }17-14=3)
Add 3 to the last term 17 three times.
a_4=17+3=20,\quad a_5=20+3=23,\quad a_6=23+3=26

20,\ 23,\ 26

(iii) \dfrac12,\dfrac32,\dfrac52,\dots

Common difference:
d=\dfrac32-\dfrac12=1\quad\left(\text{check: }\dfrac52-\dfrac32=1\right)
Add 1 to the last term \dfrac52 three times.
a_4=\dfrac52+1=\dfrac72,\quad a_5=\dfrac72+1=\dfrac92,\quad a_6=\dfrac92+1=\dfrac{11}{2}

\dfrac72,\ \dfrac92,\ \dfrac{11}{2}

(iv) 0.07,0.12,0.17,\dots

Common difference:
d=0.12-0.07=0.05\quad(\text{check: }0.17-0.12=0.05)
Add 0.05 to the last term 0.17 three times.
\begin{aligned}a_4&=0.17+0.05=0.22\\ a_5&=0.22+0.05=0.27\\ a_6&=0.27+0.05=0.32\end{aligned}

0.22,\ 0.27,\ 0.32

Find the 11th term of the arithmetic sequence 0.07,0.12,0.17,\dots

Solution

Read off the first term and find the common difference.
a_1=0.07,\qquad d=0.12-0.07=0.05
Use the formula for the nth term, a_n=a_1+(n-1)d, with n=11.
a_{11}=0.07+(11-1)(0.05)
Simplify.
a_{11}=0.07+10(0.05)=0.07+0.5=0.57

a_{11}=0.57

Book answer. a_{11}=0.57

The third term of an arithmetic sequence is 14 and the ninth term is -1. Find the first four terms of the sequence.

Solution

Write the two given terms with a_n=a_1+(n-1)d.
a_3=a_1+2d=14\quad\text{(i)},\qquad a_9=a_1+8d=-1\quad\text{(ii)}
Subtract (i) from (ii) to eliminate a_1.
6d=-1-14=-15\;\Rightarrow\;d=-\dfrac{15}{6}=-\dfrac52
Put d in (i) to get a_1.
a_1=14-2d=14-2\left(-\dfrac52\right)=14+5=19
Add d repeatedly to build the first four terms.
a_2=19-\dfrac52=\dfrac{33}{2},\quad a_3=\dfrac{33}{2}-\dfrac52=14,\quad a_4=14-\dfrac52=\dfrac{23}{2}
Check: a_3=14 as given, and a_9=19+8\left(-\dfrac52\right)=19-20=-1 as given.

19,\ \dfrac{33}{2},\ 14,\ \dfrac{23}{2}

Book answer. a_1=19,\ a_2=\dfrac{33}{2},\ a_3=14,\ a_4=\dfrac{23}{2}

Find an arithmetic sequence for a_{17}=-40 and a_{28}=-73, find a_1 and d. Write first five terms of the sequence.

Solution

Write the two given terms with a_n=a_1+(n-1)d.
a_{17}=a_1+16d=-40\quad\text{(i)},\qquad a_{28}=a_1+27d=-73\quad\text{(ii)}
Subtract (i) from (ii).
11d=-73-(-40)=-33\;\Rightarrow\;d=-3
Substitute d=-3 in (i).
a_1+16(-3)=-40\;\Rightarrow\;a_1=-40+48=8
Add d=-3 repeatedly for the first five terms.
8,\ 5,\ 2,\ -1,\ -4
Check with (ii): a_{28}=8+27(-3)=8-81=-73.

a_1=8,\ d=-3;\quad 8,\ 5,\ 2,\ -1,\ -4

Book answer. a_1=8,\ a_2=5,\ a_3=2,\ a_4=-1

The fifth term of an arithmetic sequence is 19 and 11th term is 43. Find the first term and 87th term.

Solution

Write the two given terms.
a_5=a_1+4d=19\quad\text{(i)},\qquad a_{11}=a_1+10d=43\quad\text{(ii)}
Subtract (i) from (ii).
6d=24\;\Rightarrow\;d=4
Substitute d=4 in (i).
a_1=19-4(4)=3
Now use the formula with n=87.
a_{87}=a_1+86d=3+86(4)=3+344=347

a_1=3,\quad a_{87}=347

Book answer. a_{87}=347

Which term of the sequence -6,-2,2,\dots is 70?

Solution

The sequence is arithmetic: a_1=-6 and d=-2-(-6)=4. Let the nth term be 70.
a_n=a_1+(n-1)d=70
Substitute and solve for n.
-6+(n-1)(4)=70\;\Rightarrow\;4(n-1)=76\;\Rightarrow\;n-1=19\;\Rightarrow\;n=20
n is a positive whole number, so 70 is a term of the sequence. Check: a_{20}=-6+19(4)=70.

n=20\ \text{(the 20th term)}

Book answer. a_{20}=70

Which term of the sequence \dfrac52,\dfrac32,\dfrac12,\dots is -\dfrac{105}{2}?

Solution

The sequence is arithmetic with a_1=\dfrac52 and common difference d=\dfrac32-\dfrac52=-1. Let the nth term be -\dfrac{105}{2}.
a_n=a_1+(n-1)d=-\dfrac{105}{2}
Substitute and solve for n.
\dfrac52+(n-1)(-1)=-\dfrac{105}{2}\;\Rightarrow\;-(n-1)=-\dfrac{105}{2}-\dfrac52=-\dfrac{110}{2}=-55
So n-1=55.
n=56
Check: a_{56}=\dfrac52-55=-\dfrac{105}{2}.

n=56\ \text{(the 56th term)}

Book answer. a_{56}=-\dfrac{105}{2}

If \dfrac1a,b,\dfrac1c are in A.P. Show that the common difference is \dfrac{a-c}{2ac}.

Solution

In an A.P. the common difference is the difference of consecutive terms, so d equals the middle term minus the first, and also the last minus the middle.
d=b-\dfrac1a\quad\text{(i)},\qquad d=\dfrac1c-b\quad\text{(ii)}
Add (i) and (ii). The terms b cancel.
2d=\dfrac1c-\dfrac1a
Write the right side over the common denominator ac.
2d=\dfrac{a-c}{ac}
Divide by 2.
d=\dfrac{a-c}{2ac}

d=\dfrac{a-c}{2ac}

Book answer. d=\dfrac{a-c}{2ac}

During a free fall, a sky diver falls 16 feet in the first second, 48 feet in the 2nd second and 80 feet in the third second. If he continues to fall at this rate, how many feet will he fall during the 8th second?

Solution

The distances 16,48,80,\dots form an arithmetic sequence because each is 32 more than the one before.
a_1=16,\qquad d=48-16=32
The distance in the 8th second is the 8th term.
a_8=a_1+7d=16+7(32)=16+224=240

a_8=240\,\text{ft}

Book answer. a_8=240 feet

If Rs. 1000 is saved on August 1, Rs. 3000 on August 2, Rs. 5000 on August 3 and so on. How much is saved till August 20?

Solution

The daily savings 1000,3000,5000,\dots form an arithmetic sequence. August 20 is the 20th day, so n=20.
a_1=1000,\qquad d=3000-1000=2000,\qquad n=20
The total saved till August 20 is the sum of the first 20 terms, S_n=\dfrac n2[2a_1+(n-1)d].
S_{20}=\dfrac{20}{2}[2(1000)+19(2000)]
Simplify.
S_{20}=10[2000+38000]=10(40000)=400000
Check using S_n=\dfrac n2(a_1+a_n): the amount saved on August 20 is a_{20}=1000+19(2000)=39000, so S_{20}=10(1000+39000)=400000.

S_{20}=\text{Rs. }400{,}000

Book answer. S_{20}=\text{Rs }39000

Note. The book's key prints S_{20}=\text{Rs }39000, but 39000 is only the amount saved on August 20 itself (a_{20}=1000+19(2000)=39000). The question asks how much is saved till August 20, which is the total S_{20}=\text{Rs. }400{,}000.

A gardener is making a triangular planting, with 35 plants in the first row, 31 in the second row, 27 in the third row and so on. If the pattern is consistent, how many plants will there be in the eighth row?

Solution

The numbers of plants 35,31,27,\dots form an arithmetic sequence, each row having 4 fewer plants.
a_1=35,\qquad d=31-35=-4
The eighth row is the 8th term.
a_8=a_1+7d=35+7(-4)=35-28=7

a_8=7

Book answer. a_8=7

Find A.M. between

Solution

(i) 7 and 17

The A.M. between a and b is A=\dfrac{a+b}{2}.
A=\dfrac{7+17}{2}=\dfrac{24}{2}=12

A=12

(ii) 3+3\sqrt2 and 7-3\sqrt2

Use A=\dfrac{a+b}{2}. In the sum the terms 3\sqrt2 and -3\sqrt2 cancel.
A=\dfrac{(3+3\sqrt2)+(7-3\sqrt2)}{2}=\dfrac{10}{2}=5

A=5

(iii) 7\sqrt5 and \sqrt5

Use A=\dfrac{a+b}{2} and add the like terms 7\sqrt5+\sqrt5=8\sqrt5.
A=\dfrac{7\sqrt5+\sqrt5}{2}=\dfrac{8\sqrt5}{2}=4\sqrt5

A=4\sqrt5

(iv) 2y+5 and 5y+3

Use A=\dfrac{a+b}{2} and collect the terms in y and the constants.
A=\dfrac{(2y+5)+(5y+3)}{2}=\dfrac{7y+8}{2}=\dfrac{7y}{2}+4

A=\dfrac{7y}{2}+4

Find 'b' if 10 is A.M between b and 20.

Solution

The A.M. of b and 20 is \dfrac{b+20}{2}. Set it equal to 10.
\dfrac{b+20}{2}=10
Multiply by 2 and solve.
b+20=20\;\Rightarrow\;b=0
Check: b,10,20 is 0,10,20, an A.P. with d=10.

b=0

Book answer. b=0

Find x and y if 2 and 13 are two arithmetic means between x and y.

Solution

Two A.M.s between x and y mean that x,2,13,y are in A.P. The common difference is the difference of the two means.
d=13-2=11
Going back one step from the first mean gives x.
x=2-d=2-11=-9
Going forward one step from the second mean gives y.
y=13+d=13+11=24
Check: -9,2,13,24 has a constant difference of 11.

x=-9,\quad y=24

Book answer. x=-9,\ y=24

Find the two arithmetic means between 5 and 17.

Solution

Let the means be A_1,A_2. Then 5,A_1,A_2,17 is an A.P. with a_1=5 and a_4=17.
a_4=a_1+3d=17\;\Rightarrow\;5+3d=17
Solve for d.
3d=12\;\Rightarrow\;d=4
Add d to 5 twice.
A_1=5+4=9,\qquad A_2=9+4=13

A_1=9,\quad A_2=13

Book answer. A_1=9,\ A_2=13

Find three arithmetic means between 2 and -18.

Solution

Let the means be A_1,A_2,A_3. Then 2,A_1,A_2,A_3,-18 is an A.P. with a_1=2 and a_5=-18.
a_5=a_1+4d=-18\;\Rightarrow\;2+4d=-18
Solve for d.
4d=-20\;\Rightarrow\;d=-5
Add d=-5 to 2 three times.
A_1=2-5=-3,\qquad A_2=-3-5=-8,\qquad A_3=-8-5=-13

A_1=-3,\quad A_2=-8,\quad A_3=-13

Book answer. A_1=-3,\ A_2=-8,\ A_3=-13

Exercise 4.3p. 125–126

Sum of an arithmetic series: S_n=\dfrac n2(a_1+a_n) when the last term is known, and S_n=\dfrac n2[2a_1+(n-1)d] when it is not. Missing quantities are found from a_n=a_1+(n-1)d.

Find the sum of the series 4+7+10+13+16+19+22+25.

Solution

The terms increase by the same amount, so the series is arithmetic. Read off a_1, d, the last term and the number of terms (count them: 8).
a_1=4,\quad d=7-4=3,\quad a_n=25,\quad n=8
Use S_n=\dfrac n2(a_1+a_n).
S_8=\dfrac82(4+25)=4(29)=116

S_8=116

Book answer. S_n=116

Find the sum of the series with a_1=2,\ a_n=200,\ n=100.

Solution

The first term, the last term and the number of terms are known, so use S_n=\dfrac n2(a_1+a_n).
S_{100}=\dfrac{100}{2}(2+200)
Simplify.
S_{100}=50(202)=10100

S_{100}=10100

Book answer. S_n=10100

Find the sum of the series with a_1=5,\ a_n=100,\ n=200.

Solution

Use S_n=\dfrac n2(a_1+a_n) with the given first term, last term and number of terms.
S_{200}=\dfrac{200}{2}(5+100)
Simplify.
S_{200}=100(105)=10500

S_{200}=10500

Book answer. S_n=10500

Find the sum of the series with a_1=4,\ n=15,\ d=3.

Solution

The last term is not given, so use S_n=\dfrac n2[2a_1+(n-1)d].
S_{15}=\dfrac{15}{2}[2(4)+(15-1)(3)]
Simplify the bracket: 8+14(3)=8+42=50.
S_{15}=\dfrac{15}{2}(50)=15(25)=375

S_{15}=375

Book answer. S_n=375

Find the sum of the series with a_1=50,\ n=20,\ d=-4.

Solution

Use S_n=\dfrac n2[2a_1+(n-1)d].
S_{20}=\dfrac{20}{2}[2(50)+(20-1)(-4)]
The bracket is 100-76=24.
S_{20}=10(24)=240

S_{20}=240

Book answer. S_n=240

Find the sum of the series -3+(-7)+(-11)+\dots+a_{10}.

Solution

The terms decrease by 4 each time, so a_1=-3 and d=-7-(-3)=-4. The sum is up to the 10th term, so n=10.
a_1=-3,\quad d=-4,\quad n=10
Use S_n=\dfrac n2[2a_1+(n-1)d].
S_{10}=\dfrac{10}{2}[2(-3)+(10-1)(-4)]=5[-6-36]
Simplify.
S_{10}=5(-42)=-210

S_{10}=-210

Book answer. -210

Find the sum of the series 9+11+13+15+\dots for n=12.

Solution

The terms increase by 2, so a_1=9, d=2 and n=12.
a_1=9,\quad d=11-9=2,\quad n=12
Use S_n=\dfrac n2[2a_1+(n-1)d].
S_{12}=\dfrac{12}{2}[2(9)+(12-1)(2)]=6[18+22]
Simplify.
S_{12}=6(40)=240

S_{12}=240

Book answer. S_n=240

Find the sum of the even numbers from 2 to 100.

Solution

The series 2+4+6+\dots+100 is arithmetic with a_1=2, d=2 and last term 100. Find the number of terms from a_n=a_1+(n-1)d.
100=2+(n-1)(2)\;\Rightarrow\;n-1=49\;\Rightarrow\;n=50
Use S_n=\dfrac n2(a_1+a_n).
S_{50}=\dfrac{50}{2}(2+100)=25(102)=2550

S_{50}=2550

Book answer. S_n=2550

Find the sum of the odd numbers from 1 to 99.

Solution

The series 1+3+5+\dots+99 is arithmetic with a_1=1, d=2 and last term 99. Find the number of terms.
99=1+(n-1)(2)\;\Rightarrow\;n-1=49\;\Rightarrow\;n=50
Use S_n=\dfrac n2(a_1+a_n).
S_{50}=\dfrac{50}{2}(1+99)=25(100)=2500
This agrees with the fact that the sum of the first n odd numbers is n^2=50^2=2500.

S_{50}=2500

Book answer. S_n=2500

Find the sum of all multiples of 4 that are between 14 and 523.

Solution

The multiples of 4 are 4,8,12,16,\dots. The first one greater than 14 is 16, and the last one less than 523 is 520 (since 4\times130=520 and 4\times131=524).
a_1=16,\quad d=4,\quad a_n=520
Find the number of terms.
520=16+(n-1)(4)\;\Rightarrow\;4(n-1)=504\;\Rightarrow\;n=127
Use S_n=\dfrac n2(a_1+a_n).
S_{127}=\dfrac{127}{2}(16+520)=\dfrac{127}{2}(536)=127(268)=34036

S_{127}=34036

Book answer. S_n=34036

Find S_n for the arithmetic series with a_1=3,\ a_n=-38,\ n=8.

Solution

Use S_n=\dfrac n2(a_1+a_n).
S_8=\dfrac82(3+(-38))=4(-35)=-140

S_8=-140

Book answer. S_n=-140

Find S_n for the arithmetic series with a_1=85,\ n=21,\ a_n=25.

Solution

Use S_n=\dfrac n2(a_1+a_n).
S_{21}=\dfrac{21}{2}(85+25)=\dfrac{21}{2}(110)=21(55)=1155

S_{21}=1155

Book answer. S_n=1155

Find S_n for the arithmetic series with a_1=34,\ n=9,\ a_n=2.

Solution

Use S_n=\dfrac n2(a_1+a_n).
S_9=\dfrac92(34+2)=\dfrac92(36)=9(18)=162

S_9=162

Book answer. 162

Find S_n for the arithmetic series with a_1=5,\ d=\dfrac12,\ n=13.

Solution

The last term is not given, so use S_n=\dfrac n2[2a_1+(n-1)d].
S_{13}=\dfrac{13}{2}\left[2(5)+(13-1)\left(\dfrac12\right)\right]
The bracket is 10+6=16.
S_{13}=\dfrac{13}{2}(16)=13(8)=104

S_{13}=104

Book answer. 104

Find S_n for the arithmetic series with a_1=91,\ d=-4,\ a_n=15.

Solution

The number of terms is not given. Find it from a_n=a_1+(n-1)d.
15=91+(n-1)(-4)\;\Rightarrow\;-4(n-1)=-76\;\Rightarrow\;n-1=19\;\Rightarrow\;n=20
Use S_n=\dfrac n2(a_1+a_n).
S_{20}=\dfrac{20}{2}(91+15)=10(106)=1060

S_{20}=1060

Book answer. S_n=1060

Find S_n for the arithmetic series with d=-4,\ n=9,\ a_n=27.

Solution

The first term is missing. Find it from a_n=a_1+(n-1)d.
27=a_1+(9-1)(-4)=a_1-32\;\Rightarrow\;a_1=59
Use S_n=\dfrac n2(a_1+a_n).
S_9=\dfrac92(59+27)=\dfrac92(86)=9(43)=387

S_9=387

Book answer. S_n=387

Find the sum of the arithmetic series 6+12+18+\dots+96.

Solution

The terms increase by 6: a_1=6, d=6, last term 96. Find n.
96=6+(n-1)(6)\;\Rightarrow\;6(n-1)=90\;\Rightarrow\;n=16
Use S_n=\dfrac n2(a_1+a_n).
S_{16}=\dfrac{16}{2}(6+96)=8(102)=816

S_{16}=816

Book answer. S_n=816

Find the sum of the arithmetic series 34+30+26+\dots+2.

Solution

The terms decrease by 4: a_1=34, d=-4, last term 2. Find n.
2=34+(n-1)(-4)\;\Rightarrow\;-4(n-1)=-32\;\Rightarrow\;n=9
Use S_n=\dfrac n2(a_1+a_n).
S_9=\dfrac92(34+2)=\dfrac92(36)=162

S_9=162

Book answer. S_n=162

Find the sum of the arithmetic series 10+4+(-2)+\dots+(-50).

Solution

The terms decrease by 6: a_1=10, d=4-10=-6, last term -50. Find n.
-50=10+(n-1)(-6)\;\Rightarrow\;-6(n-1)=-60\;\Rightarrow\;n=11
Use S_n=\dfrac n2(a_1+a_n).
S_{11}=\dfrac{11}{2}(10+(-50))=\dfrac{11}{2}(-40)=-220

S_{11}=-220

Book answer. S_n=-220

Find the first three terms of the arithmetic series with a_1=7,\ a_n=139,\ S_n=876.

Solution

Find n from S_n=\dfrac n2(a_1+a_n).
876=\dfrac n2(7+139)=73n\;\Rightarrow\;n=12
Find d from a_n=a_1+(n-1)d.
139=7+(12-1)d\;\Rightarrow\;11d=132\;\Rightarrow\;d=12
Build the first three terms by adding d.
a_1=7,\quad a_2=7+12=19,\quad a_3=19+12=31

7,\ 19,\ 31

Book answer. a_1=7,\ a_2=19,\ a_3=31

Find the first three terms of the arithmetic series with n=14,\ a_n=53,\ S_n=378.

Solution

Find a_1 from S_n=\dfrac n2(a_1+a_n).
378=\dfrac{14}{2}(a_1+53)=7(a_1+53)\;\Rightarrow\;a_1+53=54\;\Rightarrow\;a_1=1
Find d from a_n=a_1+(n-1)d.
53=1+13d\;\Rightarrow\;13d=52\;\Rightarrow\;d=4
Build the first three terms.
a_1=1,\quad a_2=1+4=5,\quad a_3=5+4=9

1,\ 5,\ 9

Book answer. a_1=1,\ a_2=5,\ a_3=9

Find the first three terms of the arithmetic series with a_1=6,\ a_n=306,\ S_n=1716.

Solution

Find n from S_n=\dfrac n2(a_1+a_n).
1716=\dfrac n2(6+306)=156n\;\Rightarrow\;n=11
Find d from a_n=a_1+(n-1)d.
306=6+10d\;\Rightarrow\;10d=300\;\Rightarrow\;d=30
Build the first three terms.
a_1=6,\quad a_2=6+30=36,\quad a_3=36+30=66

6,\ 36,\ 66

Book answer. a_1=6,\ a_2=36,\ a_3=66

A formation of a marching band has 14 marches in the front row, 16 in the second row, 18 in the third row and so on, for 25 rows. How many marchers are in the last row? How many marchers are there altogether?

Solution

The numbers of marchers 14,16,18,\dots form an arithmetic sequence with 2 more in each row.
a_1=14,\quad d=2,\quad n=25
The last row is the 25th term.
a_{25}=a_1+24d=14+24(2)=62
The total is the sum of the first 25 terms.
S_{25}=\dfrac{25}{2}(14+62)=\dfrac{25}{2}(76)=25(38)=950

a_{25}=62,\quad S_{25}=950

Book answer. a_{25}=62 marchers in the last row and S_{25}=950 in all.

Note. The key prints "a_{25}=62{,}950": the comma separates the two answers, 62 in the last row and 950 altogether.

How many poles will be in a pile of telephone poles if there are 50 in the first layer, 49 in the second and so on, until there are 6 in the last layer?

Solution

The layers 50,49,48,\dots,6 form an arithmetic sequence with d=-1. First find the number of layers n.
6=50+(n-1)(-1)\;\Rightarrow\;n-1=44\;\Rightarrow\;n=45
The number of poles in the pile is the sum of all the layers.
S_{45}=\dfrac{45}{2}(50+6)=\dfrac{45}{2}(56)=45(28)=1260

S_{45}=1260\ \text{poles}

Book answer. 45

Note. The key gives 45, which is the number of layers in the pile. The question asks for the number of poles in the pile, which is the sum S_{45}=1260.

A family saves money in an arithmetic sequence: Rs. 6000 in the first year, Rs. 70,000 in second year and so on, for 20 years. How much do they save in all?

Solution

Rs. 6000 and Rs. 70{,}000 are the first two terms, so the common difference is the increase from year to year.
a_1=6000,\quad d=70000-6000=64000,\quad n=20
Use S_n=\dfrac n2[2a_1+(n-1)d].
S_{20}=\dfrac{20}{2}[2(6000)+19(64000)]=10[12000+1216000]
Simplify.
S_{20}=10(1228000)=12280000

S_{20}=\text{Rs. }12{,}280{,}000

Book answer. 12{,}280{,}000

Mr. Saleem saves Rs. 500 on October 1, Rs. 550 on October 2, and Rs. 600 on October 3 and so on. How much is saved during October? (October has 31 days)

Solution

The daily savings 500,550,600,\dots form an arithmetic sequence over the 31 days of October.
a_1=500,\quad d=50,\quad n=31
Use S_n=\dfrac n2[2a_1+(n-1)d].
S_{31}=\dfrac{31}{2}[2(500)+30(50)]=\dfrac{31}{2}[1000+1500]
Simplify.
S_{31}=\dfrac{31}{2}(2500)=31(1250)=38750

S_{31}=\text{Rs. }38{,}750

Book answer. 38{,}750

Exercise 4.4p. 129–130

Geometric sequences: test for a common ratio r (the ratio of consecutive terms), use a_n=a_1r^{n-1}, and find geometric means by solving for r (or G^2=ab for a single mean).

Determine whether the sequence 5,20,100,500,\dots is geometric. If so, find the common ratio.

Solution

A sequence is geometric when the ratio of every term to the term before it is the same. Find the ratios of consecutive terms.
\dfrac{20}{5}=4,\qquad \dfrac{100}{20}=5,\qquad \dfrac{500}{100}=5
The first ratio 4 differs from the others, so there is no common ratio.

\text{Not a geometric sequence}

Book answer. The sequence is not geometric

Determine whether the sequence 2,4,6,8,\dots is geometric. If so, find the common ratio.

Solution

Find the ratios of consecutive terms.
\dfrac42=2,\qquad \dfrac64=\dfrac32,\qquad \dfrac86=\dfrac43
The ratios are not equal, so there is no common ratio. (The differences are all 2, so this is an arithmetic sequence, not a geometric one.)

\text{Not a geometric sequence}

Book answer. The sequence is not geometric

Determine whether the sequence \dfrac32,\dfrac94,\dfrac{27}{8},\dfrac{81}{16},\dots is geometric. If so, find the common ratio.

Solution

Find the ratios of consecutive terms.
\dfrac{9/4}{3/2}=\dfrac94\times\dfrac23=\dfrac32,\quad \dfrac{27/8}{9/4}=\dfrac{27}{8}\times\dfrac49=\dfrac32,\quad \dfrac{81/16}{27/8}=\dfrac{81}{16}\times\dfrac{8}{27}=\dfrac32
All the ratios are equal, so the sequence is geometric.

\text{Geometric},\quad r=\dfrac32

Book answer. The sequence is geometric (r=\dfrac32)

Determine whether the sequence 7,14,21,28,\dots is geometric. If so, find the common ratio.

Solution

Find the ratios of consecutive terms.
\dfrac{14}{7}=2,\qquad \dfrac{21}{14}=\dfrac32,\qquad \dfrac{28}{21}=\dfrac43
The ratios are not equal, so the sequence is not geometric. (It is arithmetic with d=7.)

\text{Not a geometric sequence}

Book answer. The sequence is not geometric

Find the first four terms of the geometric sequence with a_1=3,\ r=-2.

Solution

Each term of a geometric sequence is the previous term multiplied by the common ratio r.
\begin{aligned} a_1&=3 \\ a_2&=a_1r=3\times(-2)=-6 \\ a_3&=a_2r=-6\times(-2)=12 \\ a_4&=a_3r=12\times(-2)=-24 \end{aligned}

a_1=3,\ a_2=-6,\ a_3=12,\ a_4=-24

Book answer. a_1=3, a_2=-6, a_3=12, a_4=-24

Find the first four terms of the geometric sequence with a_1=27,\ r=-\dfrac13.

Solution

Each term of a geometric sequence is the previous term multiplied by the common ratio r.
\begin{aligned} a_1&=27 \\ a_2&=a_1r=27\times(-\dfrac{1}{3})=-9 \\ a_3&=a_2r=-9\times(-\dfrac{1}{3})=3 \\ a_4&=a_3r=3\times(-\dfrac{1}{3})=-1 \end{aligned}

a_1=27,\ a_2=-9,\ a_3=3,\ a_4=-1

Book answer. a_1=27, a_2=-9, a_3=3, a_4=-1

Find the first four terms of the geometric sequence with a_1=12,\ r=\dfrac12.

Solution

Each term of a geometric sequence is the previous term multiplied by the common ratio r.
\begin{aligned} a_1&=12 \\ a_2&=a_1r=12\times\dfrac{1}{2}=6 \\ a_3&=a_2r=6\times\dfrac{1}{2}=3 \\ a_4&=a_3r=3\times\dfrac{1}{2}=\dfrac{3}{2} \end{aligned}

a_1=12,\ a_2=6,\ a_3=3,\ a_4=\dfrac{3}{2}

Book answer. a_1=12, a_2=6, a_3=3, a_4=\dfrac{3}{2}

Find the next two terms of the geometric sequence 90,30,10,\dots.

Solution

Find the common ratio by dividing a term by the term before it.
r=\dfrac{a_2}{a_1}=\dfrac{30}{90}=\dfrac{1}{3}\quad\left(\text{check: }\dfrac{a_3}{a_2}=\dfrac{1}{3}\right)
Multiply the last given term by r to get the 4th term, and again for the 5th.
a_4=10\times\dfrac{1}{3}=\dfrac{10}{3},\qquad a_5=\dfrac{10}{3}\times\dfrac{1}{3}=\dfrac{10}{9}

\dfrac{10}{3},\ \dfrac{10}{9}

Book answer. a_4=\dfrac{10}{3},\ a_5=\dfrac{10}{9}

Find the next two terms of the geometric sequence 2,6,18,\dots.

Solution

Find the common ratio by dividing a term by the term before it.
r=\dfrac{a_2}{a_1}=\dfrac{6}{2}=3\quad\left(\text{check: }\dfrac{a_3}{a_2}=3\right)
Multiply the last given term by r to get the 4th term, and again for the 5th.
a_4=18\times3=54,\qquad a_5=54\times3=162

54,\ 162

Book answer. a_4=54,\ a_5=162

Find the next two terms of the geometric sequence 20,30,45,\dots.

Solution

Find the common ratio by dividing a term by the term before it.
r=\dfrac{a_2}{a_1}=\dfrac{30}{20}=\dfrac{3}{2}\quad\left(\text{check: }\dfrac{a_3}{a_2}=\dfrac{3}{2}\right)
Multiply the last given term by r to get the 4th term, and again for the 5th.
a_4=45\times\dfrac{3}{2}=\dfrac{135}{2},\qquad a_5=\dfrac{135}{2}\times\dfrac{3}{2}=\dfrac{405}{4}

\dfrac{135}{2},\ \dfrac{405}{4}

Book answer. a_4=\dfrac{135}{2},\ a_5=\dfrac{405}{4}

Find the next two terms of the geometric sequence 729,243,81,\dots.

Solution

Find the common ratio by dividing a term by the term before it.
r=\dfrac{a_2}{a_1}=\dfrac{243}{729}=\dfrac{1}{3}\quad\left(\text{check: }\dfrac{a_3}{a_2}=\dfrac{1}{3}\right)
Multiply the last given term by r to get the 4th term, and again for the 5th.
a_4=81\times\dfrac{1}{3}=27,\qquad a_5=27\times\dfrac{1}{3}=9

27,\ 9

Book answer. a_4=27,\ a_5=9

Find the next two terms of the geometric sequence \dfrac1{27},\dfrac19,\dfrac13,\dots.

Solution

Find the common ratio by dividing a term by the term before it.
r=\dfrac{a_2}{a_1}=\dfrac{\dfrac{1}{9}}{\dfrac{1}{27}}=3\quad\left(\text{check: }\dfrac{a_3}{a_2}=3\right)
Multiply the last given term by r to get the 4th term, and again for the 5th.
a_4=\dfrac{1}{3}\times3=1,\qquad a_5=1\times3=3

1,\ 3

Book answer. a_4=1,\ a_5=3

Find the next two terms of the geometric sequence \dfrac14,\dfrac12,1,\dots.

Solution

Find the common ratio by dividing a term by the term before it.
r=\dfrac{a_2}{a_1}=\dfrac{\dfrac{1}{2}}{\dfrac{1}{4}}=2\quad\left(\text{check: }\dfrac{a_3}{a_2}=2\right)
Multiply the last given term by r to get the 4th term, and again for the 5th.
a_4=1\times2=2,\qquad a_5=2\times2=4

2,\ 4

Book answer. a_4=2,\ a_5=4

Find the nth term of the geometric sequence with a_1=4,\ n=3,\ r=5.

Solution

Use a_n=a_1r^{n-1} with the given a_1, n and r.
a_{3}=4\times5^{3-1}=4\times5^{2}
Evaluate the power and multiply.
5^{2}=25\;\Rightarrow\;a_{3}=4\times25=100

a_{3}=100

Book answer. a_{3}=100

Find the nth term of the geometric sequence with a_1=2,\ n=5,\ r=2.

Solution

Use a_n=a_1r^{n-1} with the given a_1, n and r.
a_{5}=2\times2^{5-1}=2\times2^{4}
Evaluate the power and multiply.
2^{4}=16\;\Rightarrow\;a_{5}=2\times16=32

a_{5}=32

Book answer. a_{5}=32

Find the nth term of the geometric sequence with a_1=7,\ n=4,\ r=2.

Solution

Use a_n=a_1r^{n-1} with the given a_1, n and r.
a_{4}=7\times2^{4-1}=7\times2^{3}
Evaluate the power and multiply.
2^{3}=8\;\Rightarrow\;a_{4}=7\times8=56

a_{4}=56

Book answer. a_{4}=56

Find the nth term of the geometric sequence with a_1=243,\ n=5,\ r=-\dfrac{1}{3}.

Solution

Use a_n=a_1r^{n-1} with the given a_1, n and r.
a_{5}=243\times\left(-\dfrac{1}{3}\right)^{5-1}=243\times\left(-\dfrac{1}{3}\right)^{4}
Evaluate the power and multiply. A negative ratio raised to an even power is positive and to an odd power is negative.
\left(-\dfrac{1}{3}\right)^{4}=\dfrac{1}{81}\;\Rightarrow\;a_{5}=243\times\left(\dfrac{1}{81}\right)=3

a_{5}=3

Book answer. a_{5}=3

Find the nth term of the geometric sequence with a_1=32,\ n=6,\ r=-\dfrac{1}{2}.

Solution

Use a_n=a_1r^{n-1} with the given a_1, n and r.
a_{6}=32\times\left(-\dfrac{1}{2}\right)^{6-1}=32\times\left(-\dfrac{1}{2}\right)^{5}
Evaluate the power and multiply. A negative ratio raised to an even power is positive and to an odd power is negative.
\left(-\dfrac{1}{2}\right)^{5}=-\dfrac{1}{32}\;\Rightarrow\;a_{6}=32\times\left(-\dfrac{1}{32}\right)=-1

a_{6}=-1

Book answer. a_{6}=-1

Find the nth term of the geometric sequence with a_1=16,\ n=8,\ r=\dfrac{1}{2}.

Solution

Use a_n=a_1r^{n-1} with the given a_1, n and r.
a_{8}=16\times\left(\dfrac{1}{2}\right)^{8-1}=16\times\left(\dfrac{1}{2}\right)^{7}
Evaluate the power and multiply.
\left(\dfrac{1}{2}\right)^{7}=\dfrac{1}{128}\;\Rightarrow\;a_{8}=16\times\left(\dfrac{1}{128}\right)=\dfrac{1}{8}

a_{8}=\dfrac{1}{8}

Book answer. a_{8}=\dfrac{1}{8}

Find the missing geometric means: 3,\ \_\_,\ \_\_,\ \_\_,\ 48.

Solution

There are 3 means between 3 and 48, so 48 is the 5th term: a_1=3, a_5=48.
a_5=a_1r^4\;\Rightarrow\;48=3r^4
Solve for r. Take the positive root, as the book does when both end numbers are positive.
r^4=16\;\Rightarrow\;r=2
Multiply by r step by step.
a_2=3(2)=6,\qquad a_3=6(2)=12,\qquad a_4=12(2)=24
Check: a_5=24(2)=48.

6,\ 12,\ 24

Book answer. 6,12,24

Note. Since r^4=16 also allows r=-2, the sequence 3,-6,12,-24,48 is a second possibility. As in the book, the positive ratio is used.

Find the missing geometric means: 1,\ \_\_,\ \_\_,\ 8.

Solution

There are 2 means between 1 and 8, so 8 is the 4th term: a_1=1, a_4=8.
a_4=a_1r^3\;\Rightarrow\;8=1\cdot r^3
Take the cube root.
r^3=8\;\Rightarrow\;r=2
Multiply by r step by step.
a_2=1(2)=2,\qquad a_3=2(2)=4
Check: a_4=4(2)=8.

2,\ 4

Book answer. 2,4

Find the missing geometric means: 8,\ \_\_,\ \_\_,\ \_\_,\ \_\_,\ \dfrac14.

Solution

There are 4 means between 8 and \dfrac14, so \dfrac14 is the 6th term: a_1=8, a_6=\dfrac14.
a_6=a_1r^5\;\Rightarrow\;\dfrac14=8r^5
Solve for r.
r^5=\dfrac{1}{32}=\left(\dfrac12\right)^5\;\Rightarrow\;r=\dfrac12
Multiply by r step by step.
a_2=8\left(\dfrac12\right)=4,\quad a_3=2,\quad a_4=1,\quad a_5=\dfrac12
Check: a_6=\dfrac12\cdot\dfrac12=\dfrac14.

4,\ 2,\ 1,\ \dfrac12

Book answer. 4,2,1,\dfrac12

Find the missing geometric mean: 3,\ \_\_,\ 75.

Solution

There is 1 mean G between 3 and 75. Use G^2=ab (Section 4.5).
G^2=3\times75=225
Take the square root. Both end numbers are positive, so the positive root is chosen.
G=\sqrt{225}=15
Check: 3,15,75 has ratio 5 throughout.

G=15

Book answer. 15

Note. Mathematically G=-15 also satisfies G^2=225 (the sequence 3,-15,75 with r=-5). The book's Key Facts choose the positive root when both numbers are positive.

Find the missing geometric means: 5,\ \_\_,\ \_\_,\ \_\_,\ 80.

Solution

There are 3 means between 5 and 80, so 80 is the 5th term: a_1=5, a_5=80.
a_5=a_1r^4\;\Rightarrow\;80=5r^4
Solve for r (positive root).
r^4=16\;\Rightarrow\;r=2
Multiply by r step by step.
a_2=5(2)=10,\qquad a_3=20,\qquad a_4=40
Check: a_5=40(2)=80.

10,\ 20,\ 40

Book answer. 10,20,40

Note. r=-2 is also a root of r^4=16 and gives -10,20,-40. The positive ratio is used, as in the book.

Find the missing geometric means: 7,\ \_\_,\ \_\_,\ \_\_,\ 112.

Solution

There are 3 means between 7 and 112, so 112 is the 5th term: a_1=7, a_5=112.
a_5=a_1r^4\;\Rightarrow\;112=7r^4
Solve for r (positive root).
r^4=16\;\Rightarrow\;r=2
Multiply by r step by step.
a_2=7(2)=14,\qquad a_3=28,\qquad a_4=56
Check: a_5=56(2)=112.

14,\ 28,\ 56

Book answer. 14,28,56

Note. r=-2 is also a root of r^4=16 and gives -14,28,-56. The positive ratio is used, as in the book.

A Ping-Pong ball is dropped from a height of 16\,\mathrm{ft} and always rebounds one-fourth of the distance fallen. How high does it rebound the 6th time?

Solution

Each rebound height is \dfrac14 of the previous height, so the rebound heights form a geometric sequence. The first rebound is \dfrac14 of 16.
a_1=16\times\dfrac14=4,\qquad r=\dfrac14
The 6th rebound is the 6th term, a_6=a_1r^5.
a_6=4\left(\dfrac14\right)^5=\dfrac{4}{4^5}=\dfrac{1}{4^4}=\dfrac1{256}
Check by taking the drop height 16 as the starting term: 16\left(\dfrac14\right)^6=\dfrac{16}{4096}=\dfrac1{256}.

\dfrac{1}{256}\,\text{ft}

Book answer. \dfrac1{256} ft

A city has a current population of 100{,}000 and the population is increasing by 3\% each year. What will the population be in 15^{\text{th}} years?

Solution

An increase of 3\% multiplies the population by 1+0.03=1.03 every year, so the populations form a geometric sequence. The present population is the first term.
a_1=100000,\qquad r=1.03
As in the book's Example 4 ('after 15 seconds' means the 16th term), the population after 15 years is the 16th term.
a_{16}=a_1r^{15}=100000(1.03)^{15}
Evaluate with a calculator: (1.03)^{15}\approx1.55797.
a_{16}\approx100000\times1.557967=155796.7

\approx155{,}797\ \text{people}

Book answer. 151258.9 (approx.)

Note. The key's 151258.9 equals 100000(1.03)^{14}, that is the 15th term when the present population is taken as the first term. Here the population is found after 15 years, which is 100000(1.03)^{15}\approx155{,}797. If the present year is counted as year 1, the 15th year gives the key's 151{,}259.

A super ball dropped from the top of the tower (556\,\mathrm{ft} high) always rebounds three-fourths of the distance fallen. How far (up and down) will the ball have travelled when it hits the ground for the 6th time?

Solution

The ball first falls 556\,\mathrm{ft} (first time it hits the ground). Before it hits the ground the 6th time it makes 5 rebounds, and each rebound goes up and then comes down the same height.
\text{Distance}=556+2(h_1+h_2+h_3+h_4+h_5)
The rebound heights form a geometric sequence: h_1=\dfrac34(556)=417 and r=\dfrac34.
h_1=417,\quad h_2=417\left(\dfrac34\right)=312.75,\quad\dots,\quad r=\dfrac34
Sum the five heights with S_n=\dfrac{a_1(1-r^n)}{1-r} (here |r|<1).
S_5=\dfrac{417\left[1-\left(\frac34\right)^5\right]}{1-\frac34}=4(417)\left(1-\dfrac{243}{1024}\right)=1668\times\dfrac{781}{1024}\approx1272.18
Add the first drop and double the sum of the rebounds.
\text{Distance}=556+2(1272.18)=556+2544.35\approx3100.35

\approx3100\ \text{ft}

Book answer. 3100 ft (approximately)

The teaching staff of high school informs its members of school cancellation by telephone. The principal calls 2 teachers, each of whom in turn calls 2 other teachers, and so on. In order to inform the entire staff, 6 rounds of calls are made. Counting the principal, find how many people are in staff at high school?

Solution

Count the people at each stage. The principal is 1 person. Round 1 informs 2 teachers, round 2 informs 4 more, round 3 informs 8 more, and each round doubles the number of new people.
1+2+4+8+16+32+64
This is a geometric series with a_1=1, r=2, and 7 terms (the principal plus 6 rounds). Use S_n=\dfrac{a_1(r^n-1)}{r-1} since r>1.
S_7=\dfrac{1(2^7-1)}{2-1}=128-1=127

127\ \text{people}

Book answer. 127

A 5-day rain caused the river to rise. After the first day, the river rose one inch. Each day the rise in the river tripled. How much had the river risen after 5 days?

Solution

The rise on successive days is 1,3,9,27,81 inches, a geometric sequence with a_1=1 and r=3. The total rise after 5 days is the sum of the five terms.
a_1=1,\qquad r=3,\qquad n=5
Use S_n=\dfrac{a_1(r^n-1)}{r-1} since r>1.
S_5=\dfrac{1(3^5-1)}{3-1}=\dfrac{243-1}{2}=121
Check by adding: 1+3+9+27+81=121.

121\ \text{inches}

Book answer. 81

Note. The key gives 81, which is only the rise on the 5th day (a_5=1\cdot3^4=81). The question asks how much the river had risen after 5 days, which is the total S_5=121 inches.

Exercise 4.5p. 133

Geometric series: S_n=\dfrac{a_1(1-r^n)}{1-r} (or \dfrac{a_1(r^n-1)}{r-1} for |r|>1), S_n=\dfrac{a_1-a_nr}{1-r} when n is unknown, and S_\infty=\dfrac{a_1}{1-r} for |r|<1. Recurring decimals are infinite geometric series.

Find the sum of the geometric series 16+16+16+\dots to 11 terms.

Solution

Here a_1=16 and the common ratio is r=\dfrac{16}{16}=1. The formula S_n=\dfrac{a_1(1-r^n)}{1-r} cannot be used because 1-r=0 (the book's note: it applies only for r\ne1).
a_1=16,\quad r=1,\quad n=11
When r=1 every term is equal to a_1, so the sum of 11 terms is simply 11 times the first term.
S_{11}=11\times16=176

S_{11}=176

Book answer. 176

Find the sum of the geometric series 75+15+3+\dots to 10 terms.

Solution

Find the ratio and use |r|<1 in the form S_n=\dfrac{a_1(1-r^n)}{1-r}.
a_1=75,\quad r=\dfrac{15}{75}=\dfrac15,\quad n=10
Substitute.
S_{10}=\dfrac{75\left[1-\left(\frac15\right)^{10}\right]}{1-\frac15}=\dfrac{75}{\frac45}\left[1-\dfrac{1}{9765625}\right]
Simplify: \dfrac{75}{4/5}=93.75 and \dfrac{1}{9765625}\approx0.0000001, which is too small to change the answer at two decimal places.
S_{10}=93.75\times0.9999999\approx93.75

S_{10}\approx93.75

Book answer. 93.15

Note. The key's 93.15 is a misprint for 93.75. The exact value is 93.75\left(1-5^{-10}\right)=93.7499904, which is 93.75 to two decimal places.

Find the sum of the geometric series with a_1=5,\ r=3,\ n=12.

Solution

Here r=3>1, so use S_n=\dfrac{a_1(r^n-1)}{r-1}.
S_{12}=\dfrac{5(3^{12}-1)}{3-1}
Evaluate 3^{12}=531441.
S_{12}=\dfrac{5(531441-1)}{2}=\dfrac{5(531440)}{2}=5(265720)=1328600

S_{12}=1328600

Book answer. 13{,}28{,}600

Note. The key prints the number with Indian-style commas, 13{,}28{,}600=1{,}328{,}600, which agrees.

Find the sum of the geometric series with a_1=256,\ r=0.75,\ n=9.

Solution

Here |r|<1, so use S_n=\dfrac{a_1(1-r^n)}{1-r}.
S_9=\dfrac{256\left[1-(0.75)^9\right]}{1-0.75}
Evaluate (0.75)^9\approx0.07508.
S_9=\dfrac{256(1-0.07508)}{0.25}=1024(0.92492)
Multiply.
S_9\approx947.11

S_9\approx947.11

Book answer. 947.11

Find the sum of the geometric series with a_1=7,\ r=2,\ n=14.

Solution

Here r=2>1, so use S_n=\dfrac{a_1(r^n-1)}{r-1}.
S_{14}=\dfrac{7(2^{14}-1)}{2-1}
Evaluate 2^{14}=16384.
S_{14}=7(16384-1)=7(16383)=114681

S_{14}=114681

Book answer. 114681

Find the sum of the geometric series with a_1=12,\ a_5=972,\ r=-3.

Solution

The fifth term is given, so the sum is the sum of n=5 terms. Check with a_5=a_1r^4.
a_5=12(-3)^4=12(81)=972\ \checkmark
Here |r|>1, so use S_n=\dfrac{a_1(r^n-1)}{r-1} with n=5.
S_5=\dfrac{12[(-3)^5-1]}{-3-1}=\dfrac{12(-243-1)}{-4}
Simplify.
S_5=\dfrac{12(-244)}{-4}=3(244)=732
Check by adding the terms: 12-36+108-324+972=732.

S_5=732

Book answer. 732

Find the sum of the geometric series with a_1=16,\ r=-\dfrac12,\ n=10.

Solution

Here |r|<1, so use S_n=\dfrac{a_1(1-r^n)}{1-r}. Since n=10 is even, \left(-\dfrac12\right)^{10}=\dfrac1{1024}.
S_{10}=\dfrac{16\left[1-\left(-\frac12\right)^{10}\right]}{1-\left(-\frac12\right)}=\dfrac{16\left(1-\frac{1}{1024}\right)}{\frac32}
Simplify.
S_{10}=16\times\dfrac{1023}{1024}\times\dfrac23=\dfrac{1023}{96}=\dfrac{341}{32}\approx10.66

S_{10}=\dfrac{341}{32}\approx10.66

Book answer. 10.66

Find the sum of the geometric series with a_1=243,\ r=-\dfrac23,\ n=5.

Solution

Here |r|<1, so use S_n=\dfrac{a_1(1-r^n)}{1-r}. Since n=5 is odd, \left(-\dfrac23\right)^5=-\dfrac{32}{243}.
S_5=\dfrac{243\left[1-\left(-\frac23\right)^5\right]}{1+\frac23}=\dfrac{243\left(1+\frac{32}{243}\right)}{\frac53}
Simplify: 243\left(1+\dfrac{32}{243}\right)=243+32=275.
S_5=275\times\dfrac35=165

S_5=165

Book answer. 165

Find the sum of the geometric series with a_1=343,\ a_4=-1,\ r=-\dfrac17.

Solution

The last term given is a_4, so the sum is of n=4 terms. Check: a_4=343\left(-\dfrac17\right)^3=-1\ \checkmark. As n is not needed, use the form S_n=\dfrac{a_1-a_nr}{1-r} (Key Facts, Section 4.6).
S_4=\dfrac{343-(-1)\left(-\frac17\right)}{1-\left(-\frac17\right)}=\dfrac{343-\frac17}{\frac87}
Multiply the numerator and denominator by 7.
S_4=\dfrac{2401-1}{8}=\dfrac{2400}{8}=300
Check by adding the terms 343,-49,7,-1: 343-49+7-1=300.

S_4=300

Book answer. 300

Find the sum of the geometric series with a_3=\dfrac34,\ a_6=\dfrac{3}{32},\ n=6.

Solution

Divide a_6=a_1r^5 by a_3=a_1r^2 to eliminate a_1.
\dfrac{a_6}{a_3}=r^3=\dfrac{3/32}{3/4}=\dfrac{1}{8}\;\Rightarrow\;r=\dfrac12
Find a_1 from a_3=a_1r^2.
\dfrac34=a_1\left(\dfrac14\right)\;\Rightarrow\;a_1=3
Use S_n=\dfrac{a_1(1-r^n)}{1-r} with n=6.
S_6=\dfrac{3\left[1-\left(\frac12\right)^6\right]}{1-\frac12}=\dfrac{3\left(\frac{63}{64}\right)}{\frac12}=\dfrac{189}{32}

S_6=\dfrac{189}{32}\approx5.91

Book answer. 189

Note. The key prints 189, which is the numerator only. Adding the terms checks the exact value: 3+\dfrac32+\dfrac34+\dfrac38+\dfrac3{16}+\dfrac3{32}=\dfrac{189}{32}.

Find a_1 for the geometric series with S_n=244,\ r=-3,\ n=5.

Solution

Use S_n=\dfrac{a_1(r^n-1)}{r-1} and substitute.
244=\dfrac{a_1[(-3)^5-1]}{-3-1}=\dfrac{a_1(-244)}{-4}=61a_1
Solve for a_1.
a_1=\dfrac{244}{61}=4

a_1=4

Book answer. 4

Find a_1 for the geometric series with S_n=32,\ r=2,\ n=6.

Solution

Use S_n=\dfrac{a_1(r^n-1)}{r-1} and substitute.
32=\dfrac{a_1(2^6-1)}{2-1}=63a_1
Solve for a_1.
a_1=\dfrac{32}{63}\approx0.51

a_1=\dfrac{32}{63}\approx0.51

Book answer. 0.51

Find a_1 for the geometric series with a_n=324,\ r=3,\ S_n=484.

Solution

The number of terms is not given, so use S_n=\dfrac{a_1-a_nr}{1-r}.
484=\dfrac{a_1-324(3)}{1-3}=\dfrac{a_1-972}{-2}
Solve for a_1.
a_1-972=-968\;\Rightarrow\;a_1=4
Check: 4(3)^{n-1}=324\Rightarrow3^{n-1}=81\Rightarrow n=5, and S_5=\dfrac{4(3^5-1)}{2}=484.

a_1=4

Book answer. 4

Find fractional notation for the infinite geometric series

Solution

(i) 0.444\dots

Write the decimal as an infinite geometric series.
0.444\dots=0.4+0.04+0.004+\dots
The first term is a_1=0.4 and the ratio is r=\dfrac{0.04}{0.4}=0.1. Since |r|<1, use S_\infty=\dfrac{a_1}{1-r}.
S_\infty=\dfrac{0.4}{1-0.1}=\dfrac{0.4}{0.9}=\dfrac49

\dfrac49

(ii) 9.99999\dots

Split off the whole number 9 and write the rest as a geometric series.
9.99999\dots=9+0.9+0.09+0.009+\dots
This is a geometric series with a_1=9 and r=0.1. Since |r|<1:
S_\infty=\dfrac{9}{1-0.1}=\dfrac{9}{0.9}=10

(iii) 0.5555\dots

Write the decimal as an infinite geometric series.
0.5555\dots=0.5+0.05+0.005+\dots
Here a_1=0.5 and r=0.1, so |r|<1.
S_\infty=\dfrac{0.5}{1-0.1}=\dfrac{0.5}{0.9}=\dfrac59

\dfrac59

(iv) 0.6666\dots

Write the decimal as an infinite geometric series.
0.6666\dots=0.6+0.06+0.006+\dots
Here a_1=0.6 and r=0.1.
S_\infty=\dfrac{0.6}{0.9}=\dfrac23

\dfrac23

(v) 0.15151515\dots

The block 15 repeats, so group the digits in pairs.
0.15151515\dots=0.15+0.0015+0.000015+\dots
Here a_1=0.15 and r=\dfrac{0.0015}{0.15}=0.01, so |r|<1.
S_\infty=\dfrac{0.15}{1-0.01}=\dfrac{0.15}{0.99}=\dfrac{15}{99}=\dfrac{5}{33}

\dfrac{5}{33}

(vi) 0.12121212\dots

The block 12 repeats.
0.12121212\dots=0.12+0.0012+0.000012+\dots
Here a_1=0.12 and r=0.01.
S_\infty=\dfrac{0.12}{0.99}=\dfrac{12}{99}=\dfrac{4}{33}

\dfrac{4}{33}

To test its elasticity, a rubber ball is dropped into a 30\,\mathrm{ft} hollow tube that is calibrated so that the scientist can measure the height of each subsequent bounce. The scientist found that on each bounce, the ball rises to a height \dfrac25 the height of the previous bounce. How far will the ball travel before it stops bouncing?

Solution

The ball first falls 30\,\mathrm{ft}. Each bounce goes up and then down the same height, and the heights are 30\left(\dfrac25\right),\ 30\left(\dfrac25\right)^2,\dots. The rebound heights form an infinite geometric series.
h_1=30\times\dfrac25=12,\quad r=\dfrac25,\quad |r|<1
The sum of all the heights uses S_\infty=\dfrac{a_1}{1-r}.
12+4.8+1.92+\dots=\dfrac{12}{1-\frac25}=\dfrac{12}{\frac35}=20
The total distance is the first fall plus the up-and-down distance of all the bounces.
\text{Distance}=30+2(20)=70

70\ \text{ft}

Book answer. 70

A hot-air balloon rises 80\,\mathrm{ft} in the first minute of flight. If in each succeeding minutes the balloon rises only 90\% as far as in the previous minute, what will be its maximum altitude if it is allowed to rise without limit?

Solution

The rises in successive minutes are 80,\ 80(0.9),\ 80(0.9)^2,\dots, an infinite geometric series.
a_1=80,\quad r=0.9,\quad |r|<1
The maximum altitude is the sum to infinity, S_\infty=\dfrac{a_1}{1-r}.
S_\infty=\dfrac{80}{1-0.9}=\dfrac{80}{0.1}=800

The altitude after n minutes grows quickly at first, then flattens and approaches 800\,\mathrm{ft} but never passes it. (The curve joins the values for whole minutes.)

800\ \text{ft}

Book answer. 800

Exercise 4.6p. 136–137

Harmonic progressions: take the reciprocals of the terms, work in the resulting A.P. (a_n=a_1+(n-1)d), then take the reciprocal of the answer. The harmonic mean of a and b is H=\dfrac{2ab}{a+b}.

Find the indicated term of the harmonic progression \dfrac19,\dfrac1{12},\dfrac1{15},\dots : 7th term.

Solution

A sequence is a harmonic sequence when the reciprocals of its terms are in A.P., so find the A.P. of reciprocals first.
9,\ 12,\ 15,\ \dots\ \text{is an A.P.},\quad a_1=9,\ d=3
Find the 7th term of the A.P. with a_n=a_1+(n-1)d.
a_7=9+(7-1)(3)=9+18=27
The 7th term of the H.P. is the reciprocal of this.
\dfrac1{a_7}=\dfrac1{27}

\dfrac{1}{27}

Book answer. \dfrac1{27}

Find the indicated term of the harmonic progression \dfrac1{11},\dfrac19,\dfrac17,\dots : 10th term.

Solution

A sequence is a harmonic sequence when the reciprocals of its terms are in A.P., so find the A.P. of reciprocals first.
11,\ 9,\ 7,\ \dots\ \text{is an A.P.},\quad a_1=11,\ d=-2
Find the 10th term of the A.P.
a_{10}=11+(10-1)(-2)=11-18=-7
The 10th term of the H.P. is the reciprocal.
\dfrac1{a_{10}}=-\dfrac17

-\dfrac{1}{7}

Book answer. -\dfrac17

Find the indicated term of the harmonic progression \dfrac1{18},\dfrac1{13},\dfrac18,\dots : 20th term.

Solution

A sequence is a harmonic sequence when the reciprocals of its terms are in A.P., so find the A.P. of reciprocals first.
18,\ 13,\ 8,\ \dots\ \text{is an A.P.},\quad a_1=18,\ d=-5
Find the 20th term of the A.P.
a_{20}=18+(20-1)(-5)=18-95=-77
The 20th term of the H.P. is the reciprocal.
\dfrac1{a_{20}}=-\dfrac1{77}

-\dfrac{1}{77}

Book answer. -\dfrac1{77}

Find the indicated term of the harmonic progression \dfrac14,\dfrac19,\dfrac1{14},\dots : nth term.

Solution

A sequence is a harmonic sequence when the reciprocals of its terms are in A.P., so find the A.P. of reciprocals first.
4,\ 9,\ 14,\ \dots\ \text{is an A.P.},\quad a_1=4,\ d=5
Write the nth term of the A.P.
a_n=4+(n-1)(5)=5n-1
The nth term of the H.P. is the reciprocal.
\dfrac1{a_n}=\dfrac{1}{5n-1}

\dfrac{1}{5n-1}

Book answer. \dfrac{1}{5n-1}

Find the indicated term of the harmonic progression \dfrac1{27},\dfrac1{20},\dfrac1{13},\dots : nth term.

Solution

A sequence is a harmonic sequence when the reciprocals of its terms are in A.P., so find the A.P. of reciprocals first.
27,\ 20,\ 13,\ \dots\ \text{is an A.P.},\quad a_1=27,\ d=-7
Write the nth term of the A.P. and simplify.
a_n=27+(n-1)(-7)=27-7n+7=34-7n
The nth term of the H.P. is the reciprocal.
\dfrac1{a_n}=\dfrac{1}{34-7n}

\dfrac{1}{34-7n}

Book answer. \dfrac{1}{34-7n}

Find the indicated term of the harmonic progression \dfrac12,\dfrac1{2\frac12},\dfrac13,\dfrac1{3\frac12},\dots : nth term.

Solution

A sequence is a harmonic sequence when the reciprocals of its terms are in A.P., so find the A.P. of reciprocals first. The mixed numbers are 2\frac12=\dfrac52 and 3\frac12=\dfrac72.
2,\ \dfrac52,\ 3,\ \dfrac72,\ \dots\ \text{is an A.P.},\quad a_1=2,\ d=\dfrac12
Write the nth term of the A.P.
a_n=2+(n-1)\left(\dfrac12\right)=\dfrac{4+n-1}{2}=\dfrac{n+3}{2}
The nth term of the H.P. is the reciprocal.
\dfrac1{a_n}=\dfrac{2}{n+3}

\dfrac{2}{n+3}

Book answer. \dfrac{1}{\dfrac{n+3}{2}}=\dfrac{2}{n+3}

Find the 14th term of H.P. \dfrac14,\dfrac17,\dfrac1{10},\dfrac1{13},\dots

Solution

A sequence is a harmonic sequence when the reciprocals of its terms are in A.P., so find the A.P. of reciprocals first.
4,\ 7,\ 10,\ 13,\ \dots\ \text{is an A.P.},\quad a_1=4,\ d=3
Find the 14th term of the A.P.
a_{14}=4+(14-1)(3)=4+39=43
The 14th term of the H.P. is the reciprocal.
\dfrac1{a_{14}}=\dfrac1{43}

\dfrac{1}{43}

Book answer. \dfrac1{43}

7,4,1,\dots is arithmetic sequence, find the 17th term of H.P.

Solution

The H.P. is made of the reciprocals of the terms of this A.P. First find the 17th term of the A.P.
a_1=7,\quad d=4-7=-3
Use a_n=a_1+(n-1)d with n=17.
a_{17}=7+(17-1)(-3)=7-48=-41
The 17th term of the H.P. is the reciprocal of a_{17}.
\dfrac1{a_{17}}=-\dfrac1{41}

-\dfrac{1}{41}

Book answer. -\dfrac1{41}

Find the 8th term of H.P. -\dfrac12,-\dfrac15,-\dfrac18,\dots

Solution

A sequence is a harmonic sequence when the reciprocals of its terms are in A.P., so find the A.P. of reciprocals first.
-2,\ -5,\ -8,\ \dots\ \text{is an A.P.},\quad a_1=-2,\ d=-3
Find the 8th term of the A.P.
a_8=-2+(8-1)(-3)=-2-21=-23
The 8th term of the H.P. is the reciprocal.
\dfrac1{a_8}=-\dfrac1{23}

-\dfrac{1}{23}

Book answer. -\dfrac1{23}

Find H.M. between 9 and 11. Also find A, H, G and show that AH=G^2.

Solution

Use the three means with a=9 and b=11 (Section 4.7). Harmonic mean:
H=\dfrac{2ab}{a+b}=\dfrac{2(9)(11)}{9+11}=\dfrac{198}{20}=\dfrac{99}{10}
Arithmetic mean:
A=\dfrac{a+b}{2}=\dfrac{9+11}{2}=10
Geometric mean (both numbers are positive, so the positive root is taken):
G=\sqrt{ab}=\sqrt{99}=3\sqrt{11}
Left side of AH=G^2:
AH=10\times\dfrac{99}{10}=99
Right side:
G^2=\left(\sqrt{99}\right)^2=99
Both sides are 99, so AH=G^2. Also A>G>H because 10>9.95>9.9.

H=\dfrac{99}{10},\quad A=10,\quad G=\sqrt{99},\quad AH=G^2=99

Book answer. H=\dfrac{99}{10}

Find H.M. between \dfrac23 and \dfrac47.

Solution

Use H=\dfrac{2ab}{a+b} with a=\dfrac23 and b=\dfrac47. First find ab and a+b.
ab=\dfrac23\times\dfrac47=\dfrac{8}{21},\qquad a+b=\dfrac23+\dfrac47=\dfrac{14+12}{21}=\dfrac{26}{21}
Substitute.
H=\dfrac{2\left(\frac{8}{21}\right)}{\frac{26}{21}}=\dfrac{16}{26}=\dfrac{8}{13}

H=\dfrac{8}{13}

Book answer. \dfrac{8}{13}

Find four H.Ms. between \dfrac13 and \dfrac1{11}.

Solution

Four H.M.s between \dfrac13 and \dfrac1{11} mean that \dfrac13,H_1,H_2,H_3,H_4,\dfrac1{11} are in H.P. Their reciprocals are in A.P.: 3,\dots,11 with 6 terms.
a_1=3,\quad a_6=11
Find the common difference of the A.P.
a_6=a_1+5d\;\Rightarrow\;11=3+5d\;\Rightarrow\;d=\dfrac85
List the four A.P. terms between 3 and 11.
a_2=3+\dfrac85=\dfrac{23}{5},\quad a_3=\dfrac{31}{5},\quad a_4=\dfrac{39}{5},\quad a_5=\dfrac{47}{5}
The H.M.s are the reciprocals of these.
H_1=\dfrac5{23},\quad H_2=\dfrac5{31},\quad H_3=\dfrac5{39},\quad H_4=\dfrac5{47}

\dfrac{5}{23},\ \dfrac{5}{31},\ \dfrac{5}{39},\ \dfrac{5}{47}

Book answer. \dfrac{5}{23},\dfrac{5}{31},\dfrac{5}{39},\dfrac{5}{47}

Exercise 4.7p. 144–145

Sigma notation and sums of series. Use \sum k=\dfrac{n(n+1)}{2}, \sum k^2=\dfrac{n(n+1)(2n+1)}{6} and \sum k^3=\left[\dfrac{n(n+1)}{2}\right]^2 after writing the general term T_k; for arithmetico-geometric series (A.P. times G.P.) use the formulas of Sections 4.9.2 and 4.9.3.

Evaluate the sum: \displaystyle\sum_{k=1}^{5}\dfrac{1}{2k}.

Solution

Write out the terms by putting k=1,2,3,4,5.
\dfrac12+\dfrac14+\dfrac16+\dfrac18+\dfrac1{10}
Add over the common denominator 120.
\dfrac{60+30+20+15+12}{120}=\dfrac{137}{120}

\dfrac{137}{120}

Book answer. \dfrac{137}{120}

Evaluate the sum: \displaystyle\sum_{k=1}^{6}\dfrac{1}{2k+1}.

Solution

Put k=1,2,\dots,6 to get the terms.
\dfrac13+\dfrac15+\dfrac17+\dfrac19+\dfrac1{11}+\dfrac1{13}
The common denominator is 3^2\cdot5\cdot7\cdot11\cdot13=45045. Convert each term: \dfrac13=\dfrac{15015}{45045}, \dfrac15=\dfrac{9009}{45045}, and so on.
\dfrac{15015+9009+6435+5005+4095+3465}{45045}
Add the numerators.
=\dfrac{43024}{45045}

\dfrac{43024}{45045}

Book answer. \dfrac{43024}{45045}

Evaluate the sum: \displaystyle\sum_{k=0}^{5}2^k.

Solution

Note that k starts at 0. Write the terms.
2^0+2^1+2^2+2^3+2^4+2^5=1+2+4+8+16+32
Add them. (Check: this is a G.P. with a_1=1, r=2, n=6, and S_6=\dfrac{2^6-1}{2-1}=63.)
=63

Book answer. 63

Evaluate the sum: \displaystyle\sum_{k=0}^{9}\pi k.

Solution

Take the constant \pi outside the sum (\sum 3a=3\sum a rule of the book's Key Facts).
\sum_{k=0}^{9}\pi k=\pi\sum_{k=0}^{9}k=\pi(0+1+2+\dots+9)
The terms 0,1,\dots,9 form an A.P. with 9 terms after the zero. Use \sum_{k=1}^{n}k=\dfrac{n(n+1)}{2} with n=9 (the term k=0 adds nothing).
\sum_{k=1}^{9}k=\dfrac{9\times10}{2}=45
Multiply by \pi.
\sum_{k=0}^{9}\pi k=45\pi

45\pi

Book answer. 45\pi

Evaluate the sum: \displaystyle\sum_{k=1}^{8}\dfrac{k}{k+1}.

Solution

Write each term as 1 minus a fraction: \dfrac{k}{k+1}=1-\dfrac{1}{k+1}.
\sum_{k=1}^{8}\dfrac{k}{k+1}=\sum_{k=1}^{8}1-\sum_{k=1}^{8}\dfrac1{k+1}=8-\left(\dfrac12+\dfrac13+\dfrac14+\dots+\dfrac19\right)
Add the fractions over the common denominator 2520: \dfrac{1260+840+630+504+420+360+315+280}{2520}.
\dfrac12+\dfrac13+\dots+\dfrac19=\dfrac{4609}{2520}
Subtract from 8=\dfrac{20160}{2520}.
\sum_{k=1}^{8}\dfrac{k}{k+1}=\dfrac{20160-4609}{2520}=\dfrac{15551}{2520}

\dfrac{15551}{2520}

Book answer. \dfrac{15551}{2520}

Evaluate the sum: \displaystyle\sum_{k=1}^{7}(-1)^k4^{k+1}.

Solution

Write the terms. The factor (-1)^k makes the signs alternate, starting with - at k=1.
-4^2+4^3-4^4+4^5-4^6+4^7-4^8=-16+64-256+1024-4096+16384-65536
This is a G.P. with a_1=-16, r=-4 and n=7. Use S_n=\dfrac{a_1(1-r^n)}{1-r}.
S_7=\dfrac{-16\left[1-(-4)^7\right]}{1-(-4)}=\dfrac{-16(1+16384)}{5}=\dfrac{-16(16385)}{5}
Simplify: 16385\div5=3277.
S_7=-16(3277)=-52432

-52432

Book answer. -52{,}432

Evaluate the sum: \displaystyle\sum_{k=0}^{5}(k^2-2k+3).

Solution

Put k=0,1,2,3,4,5 in k^2-2k+3.
3,\ 2,\ 3,\ 6,\ 11,\ 18
Add the six values.
3+2+3+6+11+18=43
Check with the sigma rules: \sum_{k=0}^{5}k^2=55, \sum_{k=0}^{5}2k=30 and \sum_{k=0}^{5}3=3\times6=18.
55-30+18=43

Book answer. 43

Evaluate the sum: \displaystyle\sum_{k=1}^{10}\dfrac{1}{k(k+1)}.

Solution

Split each term into partial fractions.
\dfrac{1}{k(k+1)}=\dfrac1k-\dfrac1{k+1}
Write out the sum. Every term cancels with the next one (the cancelling is called telescoping).
\left(1-\dfrac12\right)+\left(\dfrac12-\dfrac13\right)+\left(\dfrac13-\dfrac14\right)+\dots+\left(\dfrac1{10}-\dfrac1{11}\right)
Only the first and the last parts are left.
1-\dfrac1{11}=\dfrac{10}{11}

\dfrac{10}{11}

Book answer. \dfrac{10}{11}

Rewrite the sum using sigma notation: \dfrac12+\dfrac23+\dfrac34+\dfrac45+\dfrac56+\dots

Solution

Look at the pattern: the numerator is 1,2,3,4,5,\dots, that is k, and the denominator is one more than the numerator, k+1. So the kth term is \dfrac{k}{k+1}, starting at k=1.
\dfrac{1}{2},\ \dfrac{2}{3},\ \dfrac{3}{4},\ \dots\ \text{is}\ \dfrac{k}{k+1}
The three dots mean the series never ends, so the upper limit is \infty.
\sum_{k=1}^{\infty}\dfrac{k}{k+1}

\sum_{k=1}^{\infty}\dfrac{k}{k+1}

Book answer. \displaystyle\sum_{k=1}^{\infty}\dfrac{k}{k+1}

Rewrite the sum using sigma notation: 3+6+9+12+15.

Solution

The terms are the multiples of 3: 3\times1,3\times2,\dots,3\times5. The kth term is 3k.
3=3(1),\ 6=3(2),\ 9=3(3),\ 12=3(4),\ 15=3(5)
There are 5 terms, so k goes from 1 to 5.
\sum_{k=1}^{5}3k

\sum_{k=1}^{5}3k

Book answer. \displaystyle\sum_{k=1}^{5}3k

Rewrite the sum using sigma notation: -2+4-8+16-32+64.

Solution

Apart from the signs, the terms are powers of 2: 2^1,2^2,2^3,2^4,2^5,2^6. So the size of the kth term is 2^k.
2,\ 4,\ 8,\ 16,\ 32,\ 64=2^1,\dots,2^6
The signs are -,+,-,+,\dots. The factor (-1)^k is -1 for odd k and +1 for even k, which matches (the first term, k=1, is negative).
(-1)^k2^k
There are 6 terms.
\sum_{k=1}^{6}(-1)^k2^k

\sum_{k=1}^{6}(-1)^k2^k

Book answer. \displaystyle\sum_{k=1}^{6}(-1)^k2^k

Rewrite the sum using sigma notation: \dfrac1{1\cdot2}+\dfrac1{2\cdot3}+\dfrac1{3\cdot4}+\dfrac1{4\cdot5}+\dots

Solution

In each denominator the first factor is k and the second factor is the next number k+1. So the kth term is \dfrac{1}{k(k+1)}, starting at k=1.
\dfrac{1}{1\cdot2},\ \dfrac{1}{2\cdot3},\ \dfrac{1}{3\cdot4},\ \dots\ \text{is}\ \dfrac{1}{k(k+1)}
The series continues for ever, so the upper limit is \infty.
\sum_{k=1}^{\infty}\dfrac{1}{k(k+1)}

\sum_{k=1}^{\infty}\dfrac{1}{k(k+1)}

Book answer. \displaystyle\sum_{k=1}^{\infty}\dfrac{1}{k(k+1)}

Prove that \displaystyle\sum_{k=1}^{n}k^3=\left[\dfrac{n(n+1)}{2}\right]^2.

Solution

Use the telescoping idea of Section 4.8.2 with m=4: \sum_{k=1}^{n}\left[k^m-(k-1)^m\right]=n^m, because all the middle terms cancel.
\sum_{k=1}^{n}\left[k^4-(k-1)^4\right]=n^4\quad\text{(i)}
Expand (k-1)^4=k^4-4k^3+6k^2-4k+1 and simplify the bracket.
k^4-(k-1)^4=4k^3-6k^2+4k-1
Put this in (i) and split the sum (the rule \sum(a+b)=\sum a+\sum b and \sum3a=3\sum a).
4\sum k^3-6\sum k^2+4\sum k-\sum1=n^4
Use \sum_{k=1}^{n}1=n, \sum k=\dfrac{n(n+1)}{2} and \sum k^2=\dfrac{n(n+1)(2n+1)}{6}, then make 4\sum k^3 the subject.
4\sum k^3=n^4+n(n+1)(2n+1)-2n(n+1)+n
Simplify: n(n+1)(2n+1)-2n(n+1)=n(n+1)(2n-1)=2n^3+n^2-n.
4\sum k^3=n^4+2n^3+n^2-n+n=n^4+2n^3+n^2=n^2(n+1)^2
Divide by 4.
\sum_{k=1}^{n}k^3=\dfrac{n^2(n+1)^2}{4}=\left[\dfrac{n(n+1)}{2}\right]^2

\sum_{k=1}^{n}k^3=\left[\dfrac{n(n+1)}{2}\right]^2

Book answer. Proved. (The question has no printed answer in the key.)

Find the sum to n terms of the series whose nth term is n+1.

Solution

Replace n by k to get the general term T_k=k+1 and sum from k=1 to n.
\sum_{k=1}^{n}T_k=\sum_{k=1}^{n}(k+1)=\sum_{k=1}^{n}k+\sum_{k=1}^{n}1
Use \sum k=\dfrac{n(n+1)}{2} and \sum1=n.
=\dfrac{n(n+1)}{2}+n=\dfrac{n(n+1)+2n}{2}=\dfrac{n^2+3n}{2}

\dfrac{n(n+3)}{2}

Book answer. \dfrac{n^2+3n}{2}

Find the sum to n terms of the series whose nth term is n^2+2n.

Solution

Put T_k=k^2+2k and sum from k=1 to n.
\sum_{k=1}^{n}T_k=\sum k^2+2\sum k
Use the standard results \sum k^2=\dfrac{n(n+1)(2n+1)}{6} and \sum k=\dfrac{n(n+1)}{2}.
=\dfrac{n(n+1)(2n+1)}{6}+n(n+1)
Take out \dfrac{n(n+1)}{6} as a common factor.
=\dfrac{n(n+1)}{6}\left[(2n+1)+6\right]=\dfrac{n(n+1)(2n+7)}{6}
Multiply out the brackets.
=\dfrac n6(2n^2+9n+7)

\dfrac{n}{6}(2n^2+9n+7)

Book answer. \dfrac{n}{6}(2n^2+9n+7)

Find the sum to n terms of the series whose nth term is 3n^2+2n+1.

Solution

Put T_k=3k^2+2k+1 and sum from k=1 to n.
\sum T_k=3\sum k^2+2\sum k+\sum1
Use the standard results.
=3\cdot\dfrac{n(n+1)(2n+1)}{6}+2\cdot\dfrac{n(n+1)}{2}+n=\dfrac{n(n+1)(2n+1)}{2}+n(n+1)+n
Take out \dfrac n2 as a common factor.
=\dfrac n2\left[(n+1)(2n+1)+2(n+1)+2\right]
Expand the bracket: (2n^2+3n+1)+(2n+2)+2=2n^2+5n+5.
=\dfrac n2(2n^2+5n+5)

\dfrac{n}{2}(2n^2+5n+5)

Book answer. \dfrac{n}{2}(2n^2+5n+5)

Sum the series up to n terms: 2^2+5^2+8^2+\dots

Solution

The numbers 2,5,8,\dots form an A.P. with a_1=2 and d=3, so the kth number is 2+(k-1)3=3k-1 and the kth term of the series is (3k-1)^2.
T_k=(3k-1)^2=9k^2-6k+1
Sum from k=1 to n with the standard results.
\sum T_k=9\sum k^2-6\sum k+\sum1=\dfrac{9n(n+1)(2n+1)}{6}-3n(n+1)+n
Take out \dfrac n2 as a common factor.
=\dfrac n2\left[3(n+1)(2n+1)-6(n+1)+2\right]
Expand the bracket: (6n^2+9n+3)-(6n+6)+2=6n^2+3n-1.
=\dfrac n2(6n^2+3n-1)
Check with n=1: the sum is 2^2=4 and \dfrac12(6+3-1)=4. With n=2: 4+25=29 and \dfrac22(24+6-1)=29.

\dfrac{n}{2}(6n^2+3n-1)

Book answer. \dfrac{n}{2}(2n^2+3n-1)

Note. The key prints \dfrac n2(2n^2+3n-1), which gives 2 for n=1, but the sum of one term is 2^2=4. The correct result is \dfrac n2(6n^2+3n-1) (checked for n=1 and n=2 above); the key has misprinted 6n^2 as 2n^2.

Sum the series up to n terms: 2^2+4^2+6^2+\dots

Solution

The kth term is (2k)^2=4k^2.
T_k=(2k)^2=4k^2
Sum from k=1 to n and use \sum k^2=\dfrac{n(n+1)(2n+1)}{6}.
\sum T_k=4\sum k^2=4\cdot\dfrac{n(n+1)(2n+1)}{6}=\dfrac{2n(n+1)(2n+1)}{3}

\dfrac{2n(n+1)(2n+1)}{3}

Book answer. \dfrac{2n(n+1)(2n+1)}{3}

Sum the series up to n terms: 1^3+3^3+5^3+\dots

Solution

The odd numbers are 2k-1, so the kth term is (2k-1)^3. Expand it.
T_k=(2k-1)^3=8k^3-12k^2+6k-1
Sum from k=1 to n with the standard results \sum k^3=\dfrac{n^2(n+1)^2}{4}, \sum k^2=\dfrac{n(n+1)(2n+1)}{6}, \sum k=\dfrac{n(n+1)}{2}.
\sum T_k=2n^2(n+1)^2-2n(n+1)(2n+1)+3n(n+1)-n
Take out the common factor n and expand the bracket.
=n\left[2n(n+1)^2-2(n+1)(2n+1)+3(n+1)-1\right]=n\left[(2n^3+4n^2+2n)-(4n^2+6n+2)+(3n+2)\right]
Collect like terms: 2n^3+(4n^2-4n^2)+(2n-6n+3n)+(-2+2)=2n^3-n.
=n(2n^3-n)=n^2(2n^2-1)
Check: n=1 gives 1, and n=2 gives 1+27=28=4(7).

n^2(2n^2-1)

Book answer. n(2n^3+2n^2+11n+1)

Note. The key's answer n(2n^3+2n^2+11n+1) gives 16 for n=1, but the sum of one term is 1^3=1, so it cannot be right. The correct sum is n^2(2n^2-1) (it gives 1 for n=1, 28 for n=2 and 153 for n=3, matching 1+27+125).

Sum the series up to n terms: 2+5+10+17+\dots to n terms

Solution

Each term is 1 more than a perfect square: 2=1^2+1, 5=2^2+1, 10=3^2+1, 17=4^2+1. So T_k=k^2+1.
T_k=k^2+1
Sum from k=1 to n.
\sum T_k=\sum k^2+\sum1=\dfrac{n(n+1)(2n+1)}{6}+n
Take out \dfrac n6 as a common factor.
=\dfrac n6\left[(n+1)(2n+1)+6\right]=\dfrac n6(2n^2+3n+1+6)
Simplify.
=\dfrac{n(2n^2+3n+7)}{6}

\dfrac{n(2n^2+3n+7)}{6}

Book answer. \dfrac{n(2n^2+3n+7)}{6}

Sum the series up to n terms: 1\times4+2\times7+3\times10+\dots

Solution

The first factors are 1,2,3,\dots, that is k. The second factors 4,7,10,\dots form an A.P. with a_1=4, d=3, so the kth is 4+(k-1)3=3k+1.
T_k=k(3k+1)=3k^2+k
Sum from k=1 to n.
\sum T_k=3\sum k^2+\sum k=\dfrac{n(n+1)(2n+1)}{2}+\dfrac{n(n+1)}{2}
Take out \dfrac{n(n+1)}{2} as a common factor.
=\dfrac{n(n+1)}{2}\left[(2n+1)+1\right]=\dfrac{n(n+1)}{2}(2n+2)
Simplify.
=n(n+1)^2

n(n+1)^2

Book answer. n(n+1)^2

Sum the series up to n terms: 1\times3\times5+3\times5\times7+5\times7\times9+\dots to n terms

Solution

In each product the factors are three consecutive odd numbers, starting at the kth odd number 2k-1. So the kth term is (2k-1)(2k+1)(2k+3). Multiply out: (2k-1)(2k+1)=4k^2-1, then multiply by (2k+3).
T_k=(4k^2-1)(2k+3)=8k^3+12k^2-2k-3
Sum from k=1 to n with the standard results.
\sum T_k=8\cdot\dfrac{n^2(n+1)^2}{4}+12\cdot\dfrac{n(n+1)(2n+1)}{6}-2\cdot\dfrac{n(n+1)}{2}-3n
Simplify each part and take out the common factor n.
=n\left[2n(n+1)^2+2(n+1)(2n+1)-(n+1)-3\right]
Expand: (2n^3+4n^2+2n)+(4n^2+6n+2)-n-1-3.
=n\left(2n^3+8n^2+7n-2\right)
Check with n=1: 1\cdot3\cdot5=15 and 1(2+8+7-2)=15.

n(2n^3+8n^2+7n-2)

Book answer. n(2n^3+8n^2+7n-2)

Sum to n terms of the following series (arithmetico-geometric series): 1+2\times2+3\times2^2+4\times2^3+\dots

Solution

Split each term into an arithmetic part and a geometric part. The numbers 1,2,3,4,\dots are in A.P. with a=1, d=1. The numbers 1,2,2^2,2^3,\dots are in G.P. with b=1 and r=2.
a=1,\quad d=1,\quad r=2
Use the book's formula for the sum of n terms (Section 4.9.2).
S_n=\dfrac{a}{1-r}+\dfrac{dr(1-r^n)}{(1-r)^2}-\dfrac{(a+nd)r^n}{1-r}
Substitute. Note that 1-r=-1.
S_n=\dfrac{1}{-1}+\dfrac{1(2)(1-2^n)}{(-1)^2}-\dfrac{(1+n)2^n}{-1}
Simplify.
S_n=-1+2-2^{n+1}+(n+1)2^n=1+2^n(n+1-2)
Collect.
S_n=2^n(n-1)+1
Check: n=1 gives 1 and n=3 gives 8(2)+1=17=1+4+12.

S_n=2^n(n-1)+1

Book answer. 2^n(n-1)+1

Sum to n terms of the following series (arithmetico-geometric series): 1+4y+7y^2+10y^3+\dots

Solution

The coefficients 1,4,7,10,\dots are in A.P. with a=1, d=3. The powers 1,y,y^2,y^3,\dots are in G.P. with b=1 and r=y.
a=1,\quad d=3,\quad r=y
Use the formula for the sum of n terms.
S_n=\dfrac{1}{1-y}+\dfrac{3y(1-y^n)}{(1-y)^2}-\dfrac{(1+3n)y^n}{1-y}
Put everything over (1-y)^2.
S_n=\dfrac{(1-y)+3y(1-y^n)-(1+3n)y^n(1-y)}{(1-y)^2}
Expand the numerator and collect: 1-y+3y-3y^{n+1}-(1+3n)y^n+(1+3n)y^{n+1}.
S_n=\dfrac{1+2y-(3n+1)y^n+(3n-2)y^{n+1}}{(1-y)^2}
Check with n=1: the numerator becomes 1+2y-4y+y^2=(1-y)^2, so S_1=1.

\dfrac{1+2y-(3n+1)y^{n}+(3n-2)y^{n+1}}{(1-y)^2}

Book answer. \dfrac{3ny^{n+1}-2y^{n+1}-3ny^n-y^n+2y+1}{(y-1)^2}

Note. The key's expression is the same as the answer here: its numerator 3ny^{n+1}-2y^{n+1}-3ny^n-y^n+2y+1 equals 1+2y-(3n+1)y^n+(3n-2)y^{n+1} and (y-1)^2=(1-y)^2.

Sum to n terms of the following series (arithmetico-geometric series): 1+\dfrac47+\dfrac{7}{7^2}+\dfrac{10}{7^3}+\dots

Solution

The numerators 1,4,7,10,\dots are in A.P. with a=1, d=3. The denominators 1,7,7^2,7^3,\dots give a G.P. 1,\dfrac17,\dfrac1{7^2},\dots with r=\dfrac17.
a=1,\quad d=3,\quad r=\dfrac17
Use the formula for the sum of n terms. Note 1-r=\dfrac67 and (1-r)^2=\dfrac{36}{49}.
S_n=\dfrac{1}{\frac67}+\dfrac{3\left(\frac17\right)\left(1-\frac1{7^n}\right)}{\frac{36}{49}}-\dfrac{(1+3n)\frac1{7^n}}{\frac67}
Simplify each fraction: \dfrac76, \dfrac{7}{12}\left(1-\dfrac1{7^n}\right) and \dfrac{7(1+3n)}{6\cdot7^n}.
S_n=\dfrac76+\dfrac7{12}-\dfrac{7}{12\cdot7^n}-\dfrac{7(1+3n)}{6\cdot7^n}
Add \dfrac76+\dfrac7{12}=\dfrac{21}{12} and combine the terms with 7^n over 12\cdot7^n.
S_n=\dfrac{21}{12}-\dfrac{7+14(1+3n)}{12\cdot7^n}=\dfrac{21\cdot7^n-21-42n}{12\cdot7^n}
Write 21\cdot7^n=3\cdot7^{n+1}.
S_n=\dfrac{3\cdot7^{n+1}-42n-21}{12\cdot7^n}
Check with n=1: \dfrac{3(49)-42-21}{12(7)}=\dfrac{84}{84}=1.

\dfrac{3\cdot7^{n+1}-42n-21}{12\cdot7^{n}}

Book answer. \dfrac{3\cdot7^{n+1}-42n-21}{12\cdot7^n}

Sum to n terms of the following series (arithmetico-geometric series): 1+\dfrac72+\dfrac{13}{4}+\dfrac{19}{8}+\dfrac{25}{16}+\dots

Solution

The numerators 1,7,13,19,25,\dots are in A.P. with a=1, d=6. The denominators 1,2,4,8,16,\dots give a G.P. 1,\dfrac12,\dfrac14,\dots with r=\dfrac12.
a=1,\quad d=6,\quad r=\dfrac12
Use the formula for the sum of n terms. Note 1-r=\dfrac12 and (1-r)^2=\dfrac14.
S_n=\dfrac{1}{\frac12}+\dfrac{6\left(\frac12\right)\left(1-\frac1{2^n}\right)}{\frac14}-\dfrac{(1+6n)\frac1{2^n}}{\frac12}
Simplify each part.
S_n=2+12\left(1-\dfrac1{2^n}\right)-\dfrac{2(1+6n)}{2^n}
Collect the terms.
S_n=14-\dfrac{12+2+12n}{2^n}=14-\dfrac{14+12n}{2^n}
Divide the numerator and denominator by 2.
S_n=14-\dfrac{6n+7}{2^{n-1}}
Check with n=1: 14-\dfrac{13}{1}=1.

14-\dfrac{6n+7}{2^{n-1}}

Book answer. 14-\dfrac{6n+7}{2^{n-1}}

Find the sum to infinity of the series: 5+\dfrac73+\dfrac99+\dfrac{11}{27}+\dots

Solution

The numerators 5,7,9,11,\dots are in A.P. with a=5, d=2. The denominators 1,3,9,27,\dots give a G.P. 1,\dfrac13,\dfrac19,\dots with r=\dfrac13. Since |r|<1 the sum to infinity exists.
a=5,\quad d=2,\quad r=\dfrac13
Use the formula for the sum to infinity.
S_\infty=\dfrac{a}{1-r}+\dfrac{dr}{(1-r)^2}
Substitute.
S_\infty=\dfrac{5}{\frac23}+\dfrac{2\cdot\frac13}{\frac49}=\dfrac{15}{2}+\dfrac{\frac23}{\frac49}=\dfrac{15}{2}+\dfrac32
Add.
S_\infty=\dfrac{18}{2}=9

S_\infty=9

Book answer. 9

Find the sum to infinity of the series: 1+\dfrac25+\dfrac3{25}+\dfrac4{125}+\dots

Solution

The numerators 1,2,3,4,\dots are in A.P. with a=1, d=1. The denominators 1,5,25,125,\dots give a G.P. 1,\dfrac15,\dfrac1{25},\dots with r=\dfrac15.
a=1,\quad d=1,\quad r=\dfrac15
Use S_\infty=\dfrac{a}{1-r}+\dfrac{dr}{(1-r)^2} (valid as |r|<1).
S_\infty=\dfrac{1}{\frac45}+\dfrac{\frac15}{\frac{16}{25}}=\dfrac54+\dfrac{5}{16}
Add over the denominator 16.
S_\infty=\dfrac{20+5}{16}=\dfrac{25}{16}

S_\infty=\dfrac{25}{16}

Book answer. \dfrac{25}{16}

Find the sum to infinity of the series: 1+4x+7x^2+10x^3+\dots

Solution

The coefficients 1,4,7,10,\dots are in A.P. with a=1, d=3; the powers 1,x,x^2,\dots are in G.P. with r=x. The sum to infinity exists for |x|<1.
a=1,\quad d=3,\quad r=x
Use S_\infty=\dfrac{a}{1-r}+\dfrac{dr}{(1-r)^2}.
S_\infty=\dfrac{1}{1-x}+\dfrac{3x}{(1-x)^2}
Combine over (1-x)^2.
S_\infty=\dfrac{(1-x)+3x}{(1-x)^2}=\dfrac{1+2x}{(1-x)^2}

S_\infty=\dfrac{2x+1}{(1-x)^2},\quad |x|<1

Book answer. \dfrac{2x+1}{(1-x)^2}

Find the sum to infinity of the series: 3+\dfrac{6}{10}+\dfrac{9}{100}+\dfrac{12}{1000}+\dots

Solution

The numerators 3,6,9,12,\dots are in A.P. with a=3, d=3. The denominators 1,10,100,1000,\dots give a G.P. with r=\dfrac1{10}.
a=3,\quad d=3,\quad r=\dfrac1{10}
Use S_\infty=\dfrac{a}{1-r}+\dfrac{dr}{(1-r)^2}.
S_\infty=\dfrac{3}{\frac9{10}}+\dfrac{\frac3{10}}{\frac{81}{100}}=\dfrac{10}{3}+\dfrac{10}{27}
Add over the denominator 27.
S_\infty=\dfrac{90+10}{27}=\dfrac{100}{27}

S_\infty=\dfrac{100}{27}

Book answer. \dfrac{100}{27}

Exercise 4.8p. 147

Real-life problems. Equal increases (a fixed amount each step) give an arithmetic sequence; a fixed percentage or multiple gives a geometric sequence. Decide which, then use a_n or S_n.

A rocket rises 20 feet in the first second, 60 feet in the 2nd second and 100 feet in the third second. If it continues at this rate, how many feet will it rise in the 20^{\text{th}} second?

Solution

The distances 20,60,100,\dots increase by the same amount, so they form an arithmetic sequence (Section 4.2).
a_1=20,\quad d=60-20=40
The distance in the 20th second is the 20th term.
a_{20}=a_1+19d=20+19(40)=20+760=780

780\ \text{ft}

Book answer. 780 ft

On the results declaration day, the school wants to invite parents as well as students. Auditorium has 21 seats in the first row and each of the other rows has one more seat than the one in front of it. There are 30 rows of seats in total. If they anticipate that 1200 people will come that day, will there be a seat for everyone? Justify your answer.

Solution

The numbers of seats 21,22,23,\dots form an arithmetic sequence, and the total number of seats is the sum of 30 rows.
a_1=21,\quad d=1,\quad n=30
Use S_n=\dfrac n2[2a_1+(n-1)d].
S_{30}=\dfrac{30}{2}[2(21)+29(1)]=15(42+29)=15(71)=1065
Compare with the number of people expected: 1065<1200.
1200-1065=135\ \text{people have no seat}

1065\ \text{seats}\lt1200\ \Rightarrow\ \text{No}

Book answer. 1065, no, because the auditorium has only 1065 seats

Note. There will not be a seat for everyone: the auditorium has 1065 seats, which is 135 fewer than the 1200 people expected.

Majid retired after 30 years of employment. If his salary was Rs. 4500 in the first year and he received as increment of Rs.820 at the end of each year of service. What was his total salary after 30 years?

Solution

The yearly salaries 4500,\ 4500+820,\ 4500+2(820),\dots form an arithmetic sequence, because the same increment is added each year.
a_1=4500,\quad d=820,\quad n=30
The total salary is the sum of the 30 yearly salaries, S_n=\dfrac n2[2a_1+(n-1)d].
S_{30}=\dfrac{30}{2}[2(4500)+29(820)]=15[9000+23780]
Simplify.
S_{30}=15(32780)=491700

S_{30}=\text{Rs. }491{,}700

Book answer. 491{,}70044

Note. The key prints 491{,}70044. The figure 491{,}700 is the answer; the trailing '44' is a misprint.

You save Rs. 1 in the first day. Then each day thereafter, save double the amount you saved the day before. Find the amount you should save in the 20^{\text{th}} day of your plan.

Solution

The daily amounts 1,2,4,8,\dots form a geometric sequence (Section 4.4) with ratio 2.
a_1=1,\quad r=2
The 20th day is the 20th term, a_n=a_1r^{n-1}.
a_{20}=1\times2^{19}=524288

a_{20}=\text{Rs. }524288

Book answer. a_{20}=524288

A vacuum pump removes 1/5 of the air from a sealed container on each stroke of its piston. What percent of the air remains after five strokes of the piston?

Solution

After each stroke 1-\dfrac15=\dfrac45 of the air that was present remains. Taking the original air as 1, the amounts form a geometric sequence 1,\dfrac45,\left(\dfrac45\right)^2,\dots (Example 19).
a_1=1,\quad r=\dfrac45
After five strokes we need the 6th term, a_6=a_1r^5 (the first term is the amount before any stroke).
a_6=\left(\dfrac45\right)^5=\dfrac{1024}{3125}=0.32768
Change to a percent.
0.32768\times100\%=32.768\%\approx32.8\%

\approx32.8\%

Book answer. a_6=0.328 or 32.8\%

Aslam borrows Rs. 20000 at 11% interest compounded annually. If he pays off the loan in full at the end of four years, how much does he pay?

Solution

With interest compounded annually the amount is multiplied by 1+0.11=1.11 each year, so the amounts 20000,\ 20000(1.11),\ 20000(1.11)^2,\dots form a geometric sequence with r=1.11.
a_1=20000,\quad r=1.11
At the end of the fourth year the amount is the 5th term (a_1 is the amount at the start).
a_5=a_1r^4=20000(1.11)^4
Evaluate (1.11)^4=1.51807041.
a_5=20000\times1.51807041=30361.4082

\approx\text{Rs. }30361.41

Book answer. 30361.4082

A property dealer estimates that a piece of land will increase its value at a rate of 10% each year. If the original value of land is Rs. 450000, what will be its value in 8 years?

Solution

An increase of 10\% multiplies the value by 1.1 each year, so the values form a geometric sequence with r=1.1 and a_1=450000 (the original value).
a_1=450000,\quad r=1.1
After 8 years the value is the 9th term.
a_9=a_1r^8=450000(1.1)^8
Evaluate (1.1)^8=2.14358881.
a_9=450000\times2.14358881\approx964615

\approx\text{Rs. }964{,}615

Book answer. 964615

A man deposits in a bank Rs. 2000 in the first year, Rs. 4000 in the second year, Rs. 8000 in the third year and so on. Find the amount he will have deposited in the bank by the fifth year.

Solution

The yearly deposits 2000,4000,8000,\dots form a geometric sequence with r=2.
a_1=2000,\quad r=\dfrac{4000}{2000}=2,\quad n=5
The total deposited by the fifth year is the sum of the first 5 terms. Use S_n=\dfrac{a_1(r^n-1)}{r-1} for r>1.
S_5=\dfrac{2000(2^5-1)}{2-1}=2000(31)=62000
Check by adding: 2000+4000+8000+16000+32000=62000.

S_5=\text{Rs. }62{,}000

Book answer. 32000

Note. The key's 32000 is only the deposit made in the fifth year (a_5=2000\times2^4=32000). The amount deposited by the fifth year is the total of the five deposits, S_5=62000.

The number of bacteria in a culture increased geometrically from 16000 to 1215000 in 5 days. Find the daily rate of increase assuming the rate to be constant.

Solution

The numbers of bacteria on successive days form a geometric sequence. As in the book's Example 37, day one is the first term, so 16000 is a_1 and the count on day 5 is a_5=1215000.
a_1=16000,\quad a_5=1215000
Use a_5=a_1r^4.
1215000=16000\,r^4\;\Rightarrow\;r^4=\dfrac{1215000}{16000}=75.9375
Take the fourth root (use a calculator, or take the square root twice: \sqrt{75.9375}\approx8.714 and \sqrt{8.714}\approx2.952).
r=\sqrt[4]{75.9375}\approx2.95
The number of bacteria is multiplied by about 2.95 every day.
\text{rate of increase}=(r-1)\times100\%\approx195\%\ \text{per day}

r\approx2.95

Book answer. r=2.95

Note. The book's key gives the common ratio r=2.95, found above; as a percentage the daily increase is (2.95-1)=195\%. If instead 16000 is taken as the count at the start (day 0) and 1215000 as the count after 5 more days, then r^5=75.9375 and r\approx2.38.

A car loan is in the amount of Rs. 600000 from the bank. Interest is 9% compounded annually and the entire amount is to be paid after 10 years. How much is to be paid back?

Solution

Interest compounded annually at 9\% multiplies the amount by 1.09 each year, a geometric sequence with r=1.09.
a_1=600000,\quad r=1.09
After 10 years the amount is the 11th term.
a_{11}=a_1r^{10}=600000(1.09)^{10}
Evaluate (1.09)^{10}=2.3673637.
a_{11}=600000\times2.3673637\approx1420418.2

\approx\text{Rs. }1{,}420{,}418

Book answer. Rs. 1420418.205

Zain bought a new car and got policy from insurance company. He will pay 5000 the first year, 6125 the second year, 7250 the third year and so on, for 10 years. How much he will pay to insurance company for vehicle?

Solution

The yearly payments 5000,6125,7250,\dots increase by the same amount, so they form an arithmetic sequence. The total payment is the sum of 10 terms.
a_1=5000,\quad d=6125-5000=1125,\quad n=10
Use S_n=\dfrac n2[2a_1+(n-1)d].
S_{10}=\dfrac{10}{2}[2(5000)+9(1125)]=5[10000+10125]
Simplify.
S_{10}=5(20125)=100625

S_{10}=\text{Rs. }100{,}625

Book answer. Rs. 100625

Naveed takes a vehicle from bank after paying down payment. He deposits Rs. 13000 in a bank in first month, Rs. 14500 in the second month, Rs. 16000 in the third month and so on. Find how much total amount he has to deposit in the bank at the end of two years.

Solution

The monthly deposits 13000,14500,16000,\dots form an arithmetic sequence. Two years is 24 months.
a_1=13000,\quad d=1500,\quad n=24
Use S_n=\dfrac n2[2a_1+(n-1)d].
S_{24}=\dfrac{24}{2}[2(13000)+23(1500)]=12[26000+34500]
Simplify.
S_{24}=12(60500)=726000

S_{24}=\text{Rs. }726{,}000

Book answer. Rs. 726000

A man borrows a loan Rs. 1000000 for leasing a car and agrees to repay with a total 20 installments. Each installment is less than the preceding by Rs. 2000. What is his first installment?

Solution

The installments decrease by the same amount each time, so they form an arithmetic sequence with d=-2000. Their total must equal the loan.
S_{20}=1000000,\quad n=20,\quad d=-2000
Use S_n=\dfrac n2[2a_1+(n-1)d].
1000000=\dfrac{20}{2}[2a_1+19(-2000)]=10(2a_1-38000)
Solve for a_1.
2a_1-38000=100000\;\Rightarrow\;2a_1=138000\;\Rightarrow\;a_1=69000
Check: the last installment is 69000-19(2000)=31000 and S_{20}=10(69000+31000)=1000000.

a_1=\text{Rs. }69{,}000

Book answer. Rs. 69000

Sara pays her first installment Rs. 8000 to insurance company for the vehicle. Each installment will increase by 5%. What total amount she will pay in 24 installments?

Solution

An increase of 5\% multiplies each installment by 1.05, so the installments form a geometric sequence. The total is the sum of 24 terms.
a_1=8000,\quad r=1.05,\quad n=24
Use S_n=\dfrac{a_1(r^n-1)}{r-1} for r>1.
S_{24}=\dfrac{8000\left[(1.05)^{24}-1\right]}{1.05-1}
Evaluate (1.05)^{24}\approx3.2250999.
S_{24}=\dfrac{8000(2.2250999)}{0.05}=160000(2.2250999)\approx356015.99

S_{24}\approx\text{Rs. }356{,}016

Book answer. Rs. 356015.99

Review Exercise 4p. 148–149

Mixed practice on all the types of sequences and series of the unit: A.P., G.P., H.P., the three means, sigma notation and the standard sums, and arithmetico-geometric series.

1 (i)

How many terms of the sequence 18,15,12,\dots are needed to give a sum of 45?

Aup to 7th
Bup to 10thAnswer
Cup to 6th
Dup to 5th

Why

The sequence is arithmetic with a_1=18 and d=-3. Put S_n=45 in S_n=\dfrac n2[2a_1+(n-1)d].
\dfrac n2[36+(n-1)(-3)]=45\;\Rightarrow\;n(39-3n)=90
Rearrange into a quadratic and factorise.
3n^2-39n+90=0\;\Rightarrow\;n^2-13n+30=0\;\Rightarrow\;(n-3)(n-10)=0
So n=3 or n=10. The terms 18+15+12 already add to 45; then the terms 9,6,3,0,-3,-6,-9 add up to zero in total, so the sum is 45 again at the 10th term. Only n=10 is among the options.
S_3=45,\qquad S_{10}=5[36-27]=45

Option (a): S_7=\dfrac72[36-18]=63\ne45. Options (c) and (d) give S_6=63 and S_5=60.

Book answer. (a)

Note. The key lists (a), but S_7=63, not 45. The sum is 45 for n=3 and for n=10, and of the options only (b), up to the 10th term, is correct.

1 (ii)

Find the 20th term from the end of A.P. 2,7,12,17,\dots,222.

A222
B132
C127Answer
D122

Why

Read the A.P. backwards: 222,217,212,\dots This is an A.P. with first term 222 and common difference -5.
a_1=222,\quad d=-5
The 20th term from the end is the 20th term of this reversed A.P.
a_{20}=222+(20-1)(-5)=222-95=127

Check from the front: 222=2+(n-1)5 gives n=45 terms, so the 20th from the end is the 26th from the start, 2+25(5)=127.

Book answer. (c)

1 (iii)

In the sequence 1,2,2,3,3,3,4,4,4,4,\dots where n consecutive terms have the value n, the 22nd term is:

A6
B7Answer
C8
D9

Why

The value n occupies n places. The places used up to the end of the block of n are the sum 1+2+\dots+n=\dfrac{n(n+1)}{2}.
\text{up to }6:\ \dfrac{6\cdot7}{2}=21\ \text{terms},\qquad \text{up to }7:\ \dfrac{7\cdot8}{2}=28\ \text{terms}
The first 21 terms end with the last 6. So the 22nd term is the first term of the block of 7s.
a_{22}=7

Option (a), 6, is the 21st term, one place too early.

Book answer. (b)

1 (iv)

If a,b,c are in A.P., then 3^a,3^b,3^c are in:

AG.P.Answer
BH.P.
CA.P.
Dnone of these

Why

Since a,b,c are in A.P., b-a=c-b=d. Find the ratios of consecutive terms of 3^a,3^b,3^c using \dfrac{3^m}{3^n}=3^{m-n}.
\dfrac{3^b}{3^a}=3^{b-a}=3^d,\qquad \dfrac{3^c}{3^b}=3^{c-b}=3^d
The two ratios are equal, so 3^a,3^b,3^c have a common ratio 3^d.
3^a,\ 3^b,\ 3^c\ \text{are in G.P.}

An A.P. in the exponents gives a G.P. in the powers: adding d to the exponent multiplies the power by 3^d.

Book answer. (a)

1 (v)

What type of sequence?

A\dfrac{4}{3^{n-2}}
B\dfrac{4}{3^{n-1}}
C\dfrac{4}{3^{n}}Answer
D\dfrac{4}{3^{n+1}}

Why

Test the ratio of consecutive terms for the sequence in option (c), a_n=\dfrac{4}{3^n}.
\dfrac{a_{n+1}}{a_n}=\dfrac{4/3^{n+1}}{4/3^{n}}=\dfrac13
The ratio is the same for every n, so the sequence is geometric (a G.P.) with a_1=\dfrac43 and r=\dfrac13.
\dfrac43,\ \dfrac49,\ \dfrac4{27},\ \dots

The question as printed has no sentence saying which one to pick. The ratio of consecutive terms is \dfrac13 in all four options, so each is a G.P.; they differ only in the first term (12, 4, \dfrac43 and \dfrac49). The key marks (c).

Book answer. (c)

Note. The wording of the question is incomplete in the book. All four options are geometric sequences with common ratio \dfrac13 (the exponent in the denominator increases by 1 each time). The book's key chooses (c), \dfrac4{3^n}, which begins \dfrac43,\dfrac49,\dots

1 (vi)

0+0.1+0.01+0.001+0.0001+\dots, the sum is:

A9
B\dfrac{10}{9}
C\dfrac{9}{10}
D\dfrac19Answer

Why

The terms 0.1,0.01,0.001,\dots (the first 0 adds nothing) form an infinite G.P. with a_1=0.1 and r=0.1. Since |r|<1 the sum is S_\infty=\dfrac{a_1}{1-r}.
S_\infty=\dfrac{0.1}{1-0.1}=\dfrac{0.1}{0.9}=\dfrac19

Option (b), \dfrac{10}{9}=1.111\dots, is the sum of 1+0.1+0.01+\dots, which starts with 1 and not 0.

Book answer. (d)

1 (vii)

Find first term of the geometric series, when S_n=30,\ n=4,\ r=-2.

A6
B-6Answer
C8
D-8

Why

Use S_n=\dfrac{a_1(1-r^n)}{1-r} with n=4 and r=-2.
30=\dfrac{a_1\left[1-(-2)^4\right]}{1-(-2)}=\dfrac{a_1(1-16)}{3}=-5a_1
Solve for a_1.
a_1=\dfrac{30}{-5}=-6

Check: -6+12-24+48=30.

Book answer. (b)

1 (viii)

The arithmetic means in the sequence -7,\_\_,\_\_,5 are:

A-3,1Answer
B3,1
C-4,2
D3,-1

Why

The sequence -7,A_1,A_2,5 is an A.P. with a_1=-7 and a_4=5.
a_4=a_1+3d\;\Rightarrow\;5=-7+3d\;\Rightarrow\;d=4
Add d=4 to the first term twice.
A_1=-7+4=-3,\qquad A_2=-3+4=1

The pair -3,1 is option (a). In option (c) the difference -4,2 is 6, which does not fit with -7 and 5.

Book answer. (a)

1 (ix)

Geometric means between 1 and \dfrac13 is:

A3
B\dfrac{1}{\sqrt3}Answer
C\sqrt3
D\dfrac13

Why

The geometric mean of a and b is G=\sqrt{ab}. Both numbers are positive, so the positive root is chosen.
G=\sqrt{1\times\dfrac13}=\sqrt{\dfrac13}=\dfrac{1}{\sqrt3}

Check: 1,\dfrac1{\sqrt3},\dfrac13 has the common ratio \dfrac{1}{\sqrt3} throughout. Option (c), \sqrt3, is the reciprocal of the answer.

Book answer. (b)

1 (x)

\displaystyle\sum_{k=1}^{3}k^2 is equal to:

A14Answer
B13
C5
D30

Why

Write out the terms and add.
\sum_{k=1}^{3}k^2=1^2+2^2+3^2=1+4+9=14

Check with \dfrac{n(n+1)(2n+1)}{6} at n=3: \dfrac{3\cdot4\cdot7}{6}=14. Option (d), 30, is \sum_{k=1}^{4}k^2.

Book answer. (a)

1 (xi)

The sigma notation for the sum -2+4-6+8 is:

A\displaystyle\sum_{k=1}^{3}(-1)^k(k+1)
B\displaystyle\sum_{k=1}^{5}(-1)^k2k
C\displaystyle\sum_{k=1}^{4}(-1)^k2kAnswer
D\displaystyle\sum_{k=2}^{5}(-1)^k2k

Why

The sizes 2,4,6,8 are 2k for k=1,2,3,4. The signs -,+,-,+ are given by (-1)^k, which is -1 for odd k.
(-1)^k2k:\quad -2,\ +4,\ -6,\ +8
There are 4 terms, so k runs from 1 to 4.
\sum_{k=1}^{4}(-1)^k2k

Option (a) gives -2+3-4. Option (b) has 5 terms, ending with -10. Option (d) starts at k=2 with +4.

Book answer. (c)

1 (xii)

\displaystyle\sum_{k=1}^{n}5 is equal to:

An
B5nAnswer
C5
D5^n

Why

The term 5 does not depend on k, so the sum adds the number 5 exactly n times.
\sum_{k=1}^{n}5=5+5+\dots+5\ (n\text{ times})=5n

This is the rule \sum_{k=1}^{n}c=cn (the book's \sum_{k=1}^{n}1=n multiplied by 5).

Book answer. (b)

1 (xiii)

The general term of the given series \dfrac1{1\cdot3}+\dfrac1{2\cdot5}+\dfrac1{3\cdot7}+\dots is:

A\dfrac{1}{n(2n+1)}Answer
B\dfrac{1}{(n+1)(2n-1)}
C\dfrac{1}{n(n+2)}
D\dfrac{1}{(n-1)(2n+1)}

Why

The first factors of the denominators are 1,2,3,\dots, that is n. The second factors 3,5,7,\dots are odd numbers increasing by 2, so 2n+1.
n=1:\ 1\cdot3,\quad n=2:\ 2\cdot5,\quad n=3:\ 3\cdot7
So the general term is \dfrac{1}{n(2n+1)}.
a_n=\dfrac{1}{n(2n+1)}

Option (d) fails at n=1 because (n-1)=0 makes the denominator zero.

Book answer. (a)

1 (xiv)

In \dfrac1{4\cdot7}+\dfrac1{7\cdot10}+\dfrac1{10\cdot13}+\dots, the nth term is:

A\dfrac{1}{(3n-1)(3n+1)}
B\dfrac{1}{(3n-2)(3n+4)}
C\dfrac{1}{(3n+1)(3n+4)}Answer
D\dfrac{1}{(n+2)(n+6)}

Why

The first factors 4,7,10,\dots form an A.P. with a_1=4, d=3, so the nth is 4+3(n-1)=3n+1. The second factors 7,10,13,\dots are the next terms of the same A.P.: 3n+4.
n=1:\ (3+1)(3+4)=4\cdot7,\quad n=2:\ 7\cdot10,\quad n=3:\ 10\cdot13
So the nth term is \dfrac{1}{(3n+1)(3n+4)}.

Option (a) at n=1 gives \dfrac{1}{2\cdot4}, and option (b) gives \dfrac{1}{1\cdot7}, neither is \dfrac1{4\cdot7}.

Book answer. (c)

1 (xv)

The sum \displaystyle\sum_{r=2}^{\infty}\dfrac{1}{r^2-1} represents:

A1
B\dfrac34Answer
C\dfrac43
D\dfrac14

Why

Factorise the denominator and split into partial fractions.
\dfrac{1}{r^2-1}=\dfrac{1}{(r-1)(r+1)}=\dfrac12\left(\dfrac1{r-1}-\dfrac1{r+1}\right)
Write out the first few terms from r=2. Most terms cancel (telescoping).
\dfrac12\left[\left(1-\dfrac13\right)+\left(\dfrac12-\dfrac14\right)+\left(\dfrac13-\dfrac15\right)+\left(\dfrac14-\dfrac16\right)+\dots\right]
Only 1 and \dfrac12 remain, because every later term is cancelled by the term two places on (and the tail tends to 0).
\dfrac12\left(1+\dfrac12\right)=\dfrac12\cdot\dfrac32=\dfrac34

Check with partial sums: \dfrac13+\dfrac18+\dfrac1{15}+\dfrac1{24}\approx0.60, increasing towards 0.75.

Book answer. (b)

1 (xvi)2025 Annual I, Set B, MCQ 8

Sum of the series 1+3+5+7+9+11+\dots is:

An
Bn^2Answer
Cn(n+1)
D(n-1)

Why

The odd numbers form an A.P. with a_1=1 and d=2. For n terms use S_n=\dfrac n2[2a_1+(n-1)d].
S_n=\dfrac n2[2+2(n-1)]=\dfrac n2(2n)=n^2

Check: 1+3+5=9=3^2 and 1+3+5+7=16=4^2.

Book answer. (b)

1 (xvii)

\displaystyle\sum_{k=1}^{10}3 is equal to:

A10
B103
C300
D30Answer

Why

The term 3 does not depend on k, so it is added 10 times.
\sum_{k=1}^{10}3=3\times10=30

Option (a), 10, would be \sum_{k=1}^{10}1.

Book answer. (d)

The sum of four numbers in A.P. is 24 and their product is 945. Find the numbers.

Solution

Take the four numbers as a-3d,\ a-d,\ a+d,\ a+3d. They are in A.P. with common difference 2d, and the terms are placed so that the sum is simple.
a-3d,\quad a-d,\quad a+d,\quad a+3d
The sum is 24. The d terms cancel.
4a=24\;\Rightarrow\;a=6
The product is 945. Multiply the outer pair and the inner pair, using (x-y)(x+y)=x^2-y^2.
(a^2-9d^2)(a^2-d^2)=945\;\Rightarrow\;(36-9d^2)(36-d^2)=945
Let t=d^2 and expand.
1296-360t+9t^2=945\;\Rightarrow\;9t^2-360t+351=0\;\Rightarrow\;t^2-40t+39=0
Factorise: (t-1)(t-39)=0. Take t=1, so d=\pm1 (both signs give the same four numbers in opposite order).
d^2=1\;\Rightarrow\;d=1:\quad 6-3,\ 6-1,\ 6+1,\ 6+3
Check: sum 3+5+7+9=24 and product 3\times5\times7\times9=945.

3,\ 5,\ 7,\ 9

Book answer. 3,5,7,9

Note. The other root t=39 gives the irrational numbers 6\pm3\sqrt{39} and 6\pm\sqrt{39}, which also have sum 24 and product 945. The book expects the whole-number answer 3,5,7,9.

Find four numbers in A.P., whose sum is 6 and sum of whose square is 14.

Solution

Take the four numbers as a-3d,\ a-d,\ a+d,\ a+3d.
S=4a=6\;\Rightarrow\;a=\dfrac32
Write the sum of the squares. The cross terms cancel in pairs.
(a-3d)^2+(a-d)^2+(a+d)^2+(a+3d)^2=4a^2+20d^2=14
Put a=\dfrac32.
4\left(\dfrac94\right)+20d^2=14\;\Rightarrow\;9+20d^2=14\;\Rightarrow\;d^2=\dfrac14\;\Rightarrow\;d=\dfrac12
The numbers (taking d=\dfrac12) are:
\dfrac32-\dfrac32=0,\quad \dfrac32-\dfrac12=1,\quad \dfrac32+\dfrac12=2,\quad \dfrac32+\dfrac32=3
Check: 0+1+2+3=6 and 0+1+4+9=14.

0,\ 1,\ 2,\ 3

Book answer. 0,1,2,3

Insert 20 A.Ms. between 2 and 86.

Solution

With 20 means between 2 and 86, the A.P. 2,A_1,\dots,A_{20},86 has 22 terms, so 86 is the 22nd term.
a_{22}=a_1+21d\;\Rightarrow\;86=2+21d
Solve for d.
21d=84\;\Rightarrow\;d=4
Add d=4 repeatedly to 2.
A_1=6,\quad A_2=10,\quad A_3=14,\quad\dots,\quad A_{20}=2+20(4)=82
Check: 82+4=86.

6,\ 10,\ 14,\ \dots,\ 82

Book answer. 6,10,14,\dots

Evaluate 3+33+333+\dots up to n terms.

Solution

Take the common factor 3 out and make each number a multiple of 9 so that it becomes 10^k-1 (for example 9=10-1, 99=100-1).
3+33+333+\dots=\dfrac39\left(9+99+999+\dots\right)=\dfrac13\left(9+99+999+\dots\right)
Write each term as a power of 10 minus 1.
=\dfrac13\left[(10-1)+(10^2-1)+(10^3-1)+\dots+(10^n-1)\right]
Separate into a G.P. and a constant sum. The G.P. 10+10^2+\dots+10^n has a_1=10, r=10, so S_n=\dfrac{10(10^n-1)}{10-1}; the -1 appears n times.
=\dfrac13\left[\dfrac{10(10^n-1)}{9}-n\right]
Put over 9 and simplify.
=\dfrac13\cdot\dfrac{10^{n+1}-10-9n}{9}=\dfrac{1}{27}\left[10^{n+1}-9n-10\right]
Check with n=2: 3+33=36 and \dfrac{1}{27}(1000-18-10)=\dfrac{972}{27}=36.

\dfrac{1}{27}\left[10^{n+1}-9n-10\right]

Book answer. \dfrac{1}{27}\left[10^{n+1}-9n-10\right]

If the product of three numbers in G.P be 216 and their sum be 19, then find the numbers.

Solution

Take the three numbers as \dfrac ar,\ a,\ ar, which are in G.P. with ratio r. Their product is a^3.
\dfrac ar\cdot a\cdot ar=a^3=216\;\Rightarrow\;a=6
Their sum is 19.
\dfrac6r+6+6r=19\;\Rightarrow\;\dfrac6r+6r=13
Multiply by r and form a quadratic.
6+6r^2=13r\;\Rightarrow\;6r^2-13r+6=0
Factorise: 6r^2-13r+6=(2r-3)(3r-2).
r=\dfrac32\quad\text{or}\quad r=\dfrac23
For r=\dfrac32 the numbers are \dfrac{6}{3/2},\ 6,\ 6\left(\dfrac32\right). For r=\dfrac23 the same numbers appear in reverse order.
4,\ 6,\ 9\quad\text{(or }9,\ 6,\ 4\text{)}
Check: 4\times6\times9=216 and 4+6+9=19.

4,\ 6,\ 9

Book answer. 4,6,9

Insert 4 G.Ms. between 2 and 486.

Solution

The G.P. 2,G_1,G_2,G_3,G_4,486 has 6 terms with a_1=2 and a_6=486.
a_6=a_1r^5\;\Rightarrow\;486=2r^5
Solve for r.
r^5=243=3^5\;\Rightarrow\;r=3
Multiply by r=3 repeatedly.
G_1=2(3)=6,\quad G_2=18,\quad G_3=54,\quad G_4=162
Check: 162\times3=486.

6,\ 18,\ 54,\ 162

Book answer. 6,18,54,162

Find n so that \dfrac{a^{n+1}+b^{n+1}}{a^n+b^n} may be the harmonic means between a and b.

Solution

The harmonic mean of a and b is H=\dfrac{2ab}{a+b}. Equate it to the given expression.
\dfrac{a^{n+1}+b^{n+1}}{a^n+b^n}=\dfrac{2ab}{a+b}
Cross-multiply.
(a^{n+1}+b^{n+1})(a+b)=2ab(a^n+b^n)
Expand both sides.
a^{n+2}+a^{n+1}b+ab^{n+1}+b^{n+2}=2a^{n+1}b+2ab^{n+1}
Move everything to the left and combine: a^{n+1}b+ab^{n+1}-2a^{n+1}b-2ab^{n+1}=-a^{n+1}b-ab^{n+1}.
a^{n+2}-a^{n+1}b-ab^{n+1}+b^{n+2}=0
Factorise by grouping the first two terms and the last two terms.
a^{n+1}(a-b)-b^{n+1}(a-b)=0\;\Rightarrow\;(a-b)\left(a^{n+1}-b^{n+1}\right)=0
For two different numbers a\ne b the factor a-b is not zero, so a^{n+1}=b^{n+1}, that is \left(\dfrac ab\right)^{n+1}=1. A power of a number other than \pm1 is 1 only when the exponent is 0.
n+1=0\;\Rightarrow\;n=-1
Check with n=-1: \dfrac{a^0+b^0}{a^{-1}+b^{-1}}=\dfrac{2}{\frac{a+b}{ab}}=\dfrac{2ab}{a+b}, the harmonic mean.

n=-1

Book answer. n=-1

Find the H.P., whose 3rd and 14th terms are \dfrac67 and \dfrac13 respectively.

Solution

The reciprocals of the terms of an H.P. are in A.P. The 3rd and 14th terms of the A.P. are the reciprocals of the given terms.
a_3=\dfrac76,\qquad a_{14}=3
Write them with a_n=a_1+(n-1)d.
a_1+2d=\dfrac76\quad\text{(i)},\qquad a_1+13d=3\quad\text{(ii)}
Subtract (i) from (ii).
11d=3-\dfrac76=\dfrac{11}{6}\;\Rightarrow\;d=\dfrac16
Put d in (i).
a_1=\dfrac76-\dfrac26=\dfrac56
The A.P. is \dfrac56,\ 1,\ \dfrac76,\ \dfrac43,\dots. Take reciprocals to get the H.P.
\dfrac65,\ 1,\ \dfrac67,\ \dfrac34,\ \dots
Check: the general term of the A.P. is \dfrac56+(n-1)\dfrac16=\dfrac{n+4}{6}, so the H.P. has nth term \dfrac{6}{n+4}; n=3 gives \dfrac67 and n=14 gives \dfrac13.

\dfrac65,\ 1,\ \dfrac67,\ \dfrac34,\ \dots\quad\left(a_n=\dfrac{6}{n+4}\right)

Book answer. a_1=\dfrac65,\ a_2=1,\ a_3=\dfrac67,\ a_4=\dfrac34

Evaluate the sum:

Solution

(i) \displaystyle\sum_{k=1}^{10}\dfrac{2^k}{2^{k+1}}

Simplify each term using 2^{k+1}=2\cdot2^k.
\dfrac{2^k}{2^{k+1}}=\dfrac{2^k}{2\cdot2^k}=\dfrac12
Every one of the 10 terms is \dfrac12, so add \dfrac12 ten times.
\sum_{k=1}^{10}\dfrac12=10\times\dfrac12=5

(ii) \displaystyle\sum_{k=1}^{8}(-1)^{k+1}3^k

Write the terms. The sign is + for odd k because (-1)^{k+1}=+1 when k+1 is even.
3-9+27-81+243-729+2187-6561
This is a G.P. with a_1=3, r=-3 and n=8. Use S_n=\dfrac{a_1(1-r^n)}{1-r}.
S_8=\dfrac{3\left[1-(-3)^8\right]}{1-(-3)}=\dfrac{3(1-6561)}{4}=\dfrac{3(-6560)}{4}
Simplify.
S_8=-4920

-4920

Sum to n terms of the following (arithmetico-geometric series): 4+14+30+52+82+\dots

Solution

Look at the differences between consecutive terms.
14-4=10,\qquad 30-14=16,\qquad 52-30=22
The differences 10,16,22 increase by the constant 6, so the kth term is a quadratic T_k=\alpha k^2+\beta k+\gamma with 2\alpha=6.
\alpha=3
Use T_2-T_1=3\alpha+\beta=10 and T_1=\alpha+\beta+\gamma=4.
\beta=10-9=1,\qquad \gamma=4-3-1=0\;\Rightarrow\;T_k=3k^2+k=k(3k+1)
Check: T_3=30, T_4=52 and T_5=3(25)+5=80.
Sum from k=1 to n using \sum k^2=\dfrac{n(n+1)(2n+1)}{6} and \sum k=\dfrac{n(n+1)}{2}.
\sum T_k=3\sum k^2+\sum k=\dfrac{n(n+1)(2n+1)}{2}+\dfrac{n(n+1)}{2}
Take out \dfrac{n(n+1)}{2}.
=\dfrac{n(n+1)}{2}(2n+1+1)=n(n+1)^2

n(n+1)^2

Book answer. n(n+1)^2

Note. The fifth term printed in the book is 82, but the pattern of the first four terms gives T_5=80 and the book's own answer n(n+1)^2 gives S_5-S_4=180-100=80, so 82 is a misprint for 80. Also, this series is not arithmetico-geometric: it comes from a quadratic general term.

Sum to n terms of the following (arithmetico-geometric series): 1+4+10+21+39+\dots

Solution

Look at the differences between consecutive terms.
4-1=3,\quad 10-4=6,\quad 21-10=11,\quad 39-21=18
These differences are 1^2+2,\ 2^2+2,\ 3^2+2,\ 4^2+2. So T_{k+1}-T_k=k^2+2.
T_{k+1}=T_k+k^2+2
Add up the differences from 1 to n-1 to get the nth term T_n.
T_n=1+\sum_{k=1}^{n-1}(k^2+2)=1+\dfrac{(n-1)n(2n-1)}{6}+2(n-1)
Simplify over 6: (n-1)n(2n-1)=2n^3-3n^2+n.
T_n=\dfrac{2n^3-3n^2+n+12n-6}{6}=\dfrac{2n^3-3n^2+13n-6}{6}
Check: T_1=\dfrac{6}{6}=1, T_2=\dfrac{24}{6}=4, T_5=\dfrac{234}{6}=39.
Now sum T_k from k=1 to n with the standard results.
S_n=\dfrac16\left[2\sum k^3-3\sum k^2+13\sum k-6n\right]=\dfrac16\left[\dfrac{n^2(n+1)^2}{2}-\dfrac{n(n+1)(2n+1)}{2}+\dfrac{13n(n+1)}{2}-6n\right]
Take out \dfrac{n}{12} and expand the bracket: n(n+1)^2-(n+1)(2n+1)+13(n+1)-12.
S_n=\dfrac{n}{12}\left[(n^3+2n^2+n)-(2n^2+3n+1)+(13n+13)-12\right]=\dfrac{n}{12}\left(n^3+11n\right)
Simplify.
S_n=\dfrac{n^2(n^2+11)}{12}
Check: n=2 gives \dfrac{4(15)}{12}=5=1+4, and n=3 gives \dfrac{9(20)}{12}=15=1+4+10.

S_n=\dfrac{n^2(n^2+11)}{12}

Book answer. \dfrac16\left(2n^3-3n^2+13n-6\right)

Note. The book's answer \dfrac16(2n^3-3n^2+13n-6) is the nth term T_n (found in the working above), not the sum of n terms. The sum asked for is S_n=\dfrac{n^2(n^2+11)}{12}.

FBISE 2025 and 2026 questions

Questions from the FBISE HSSC-I 2026 (1st Annual) and FBISE HSSC-I 2025 (1st Annual) papers that belong to this unit, answered the way an examiner expects.

2025 Annual I, Set B, MCQ 51 mark

In an arithmetic sequence, a_1=5 and a_{15}=75. What is the common difference?

A3
B4
C5Answer
D6

Why

a_{15}=a_1+14d, so 75=5+14d, 14d=70 and d=5.

There are 14 steps from the 1st term to the 15th. Dividing by 15 instead of 14 gives \dfrac{70}{15}, which is not an option.

2025 Annual I, Set B, MCQ 61 mark

If \dfrac{1}{2},\dfrac{1}{x},\dfrac{1}{5} are in harmonic sequence then what is the value of x?

A\dfrac{2}{7}
B\dfrac{7}{2}Answer
C\dfrac{7}{10}
D\dfrac{10}{7}

Why

The reciprocals of an H.P. are in A.P., so 2,\ x,\ 5 are in A.P. The middle term is the arithmetic mean: x=\dfrac{2+5}{2}=\dfrac72.

Option A, \dfrac27, is \dfrac1x: it is the middle term of the H.P., not the value of x.

2025 Annual I, Set B, MCQ 71 mark

What is the first term of an infinite geometric series whose sum is 10, and common ratio 0.6?

A4Answer
B5
C6
D7

Why

S_\infty=\dfrac{a_1}{1-r}, so a_1=S_\infty(1-r)=10(1-0.6)=10(0.4)=4.

Multiplying 10 by 0.6 gives 6 (option C); that is S_\infty r, not S_\infty(1-r).

2025 Annual I, Set B, MCQ 81 mark

What is the sum of the series up to 4 terms? 1+3+5+\cdots+(2n-1)=n^2 \;\forall n\in N

A7
B9
C15
D16Answer

Why

Put n=4 in the given result: S_4=4^2=16. Direct addition agrees: 1+3+5+7=16.

Option B, 9, is the sum of 3 terms (n=3). Option C, 15, is 16-1, what is left if the +1 is dropped; the sum of 4 terms is exactly 4^2.

2026 Annual I, Set B, MCQ 101 mark

Formula of n^{th} term of an A.P is:

Aa+nd
Ba+(n-1)dAnswer
Ca+(n+1)d
Da(n-1)d

Why

The first term is a_1=a, the second is a+d, the third is a+2d. The common difference has been added n-1 times by the nth term, so a_n=a+(n-1)d.

Option A, a+nd, would give the first term as a+d, so it is really the (n+1)th term. Option D multiplies instead of adding.

2026 Annual I, Set B, MCQ 111 mark

H.P decreases when A.P:

ADecreases
BIncreasesAnswer
CConstant
DZero

Why

The terms of an H.P. are the reciprocals of the terms of an A.P. For positive terms, a larger number has a smaller reciprocal. If the A.P. increases, for example 1,2,3,\dots, the H.P. 1,\dfrac12,\dfrac13,\dots decreases.

Option A is the opposite: a decreasing A.P. such as 3,2,1 gives an increasing H.P. \dfrac13,\dfrac12,1. Option C gives a constant H.P., and option D is impossible because \dfrac10 is undefined.

2026 Annual I, Set B, Q.2 (i)4 marks

Find the sum of the geometric series \dfrac23,-1,\dfrac32,\dots up to 5 terms.

Solution

Find the common ratio by dividing a term by the one before it.
a_1=\dfrac23,\qquad r=\dfrac{-1}{2/3}=-\dfrac32
Use S_n=\dfrac{a_1(1-r^n)}{1-r} with n=5.
S_5=\dfrac{\frac23\left[1-\left(-\frac32\right)^5\right]}{1-\left(-\frac32\right)}
Evaluate \left(-\dfrac32\right)^5=-\dfrac{243}{32}, so 1-r^5=1+\dfrac{243}{32}=\dfrac{275}{32}, and 1-r=\dfrac52.
S_5=\dfrac23\cdot\dfrac{275}{32}\cdot\dfrac25=\dfrac{1100}{480}=\dfrac{55}{24}
Check by adding the five terms \dfrac23,-1,\dfrac32,-\dfrac94,\dfrac{27}{8} over the denominator 24.
\dfrac{16-24+36-54+81}{24}=\dfrac{55}{24}

S_5=\dfrac{55}{24}

2025 Annual I, Set B, Q.2 (iv) OR4 marks

A harmonic sequence has 2nd term \dfrac{1}{6} and 4th term \dfrac{1}{12}. Find general term of the sequence.

Solution

The reciprocals of an H.P. are in A.P., so the A.P. has a_2=6 and a_4=12.
a_1+d=6\quad\text{(i)},\qquad a_1+3d=12\quad\text{(ii)}
Subtract (i) from (ii).
2d=6\;\Rightarrow\;d=3
Put d=3 in (i).
a_1=6-3=3
General term of the A.P.
a_n=3+(n-1)3=3n
The general term of the H.P. is the reciprocal.
\dfrac{1}{a_n}=\dfrac{1}{3n}
Check: n=2 gives \dfrac16 and n=4 gives \dfrac1{12}.

\dfrac{1}{3n}

2025 Annual I, Set B, Q.2 (vi) (a)4 for (a) and (b) together = 4 marks

In an arithmetic sequence, the 5th term is 15, and the 12th term is 50. Find the common difference.

Solution

Write the two terms with a_n=a_1+(n-1)d.
a_1+4d=15\quad\text{(i)},\qquad a_1+11d=50\quad\text{(ii)}
Subtract (i) from (ii).
7d=35\;\Rightarrow\;d=5

d=5

2025 Annual I, Set B, Q.2 (vi) (b)4 for (a) and (b) together = 4 marks

In an arithmetic sequence, the 5th term is 15, and the 12th term is 50. Find the first term of the sequence.

Solution

From part (a), d=5. Put it in a_1+4d=15.
a_1+4(5)=15
Solve.
a_1=15-20=-5
Check: a_{12}=-5+11(5)=50.

a_1=-5

2025 Annual I, Set B, Q.2 (viii)4 marks

A bouncing ball rebounds 80% of its previous height. If the ball is dropped from a height of 50 meters. Find the total distance traveled by the ball before coming to rest.

Solution

The ball first falls 50 m. After that, each bounce goes up and then down the same height. The rebound heights are 80% of the previous one, so they form a G.P.
h_1=50(0.8)=40,\qquad r=0.8,\qquad |r|<1
Add all the rebound heights with S_\infty=\dfrac{a_1}{1-r}.
40+32+25.6+\dots=\dfrac{40}{1-0.8}=\dfrac{40}{0.2}=200
Each rebound height is travelled twice (up and down), and the first fall of 50 m is added once.
\text{Distance}=50+2(200)=450

450\,\mathrm{m}

2026 Annual I, Set B, Q.2 (ix) OR4 marks

Find the sum to n terms of the sequence: 3,9,27,81,\dots

Solution

Each term is 3 times the one before, so the sequence is geometric.
a_1=3,\qquad r=\dfrac93=3
Since r>1, use the form S_n=\dfrac{a_1(r^n-1)}{r-1}.
S_n=\dfrac{3(3^n-1)}{3-1}
Simplify.
S_n=\dfrac{3(3^n-1)}{2}
Check: n=3 gives \dfrac{3(26)}{2}=39=3+9+27.

S_n=\dfrac{3\left(3^n-1\right)}{2}

2026 Annual I, Set B, Q.2 (xi)4 marks

Find the sum of first 20 terms of A.P with a=1, d=3.

Solution

Use S_n=\dfrac n2\left[2a_1+(n-1)d\right] with n=20.
S_{20}=\dfrac{20}{2}\left[2(1)+19(3)\right]
Simplify.
S_{20}=10(2+57)=10(59)=590
Check: the 20th term is 1+19(3)=58, and S_{20}=\dfrac{20}{2}(1+58)=590.

S_{20}=590

2026 Annual I, Set B, Q.3 (a)4+2+2 = 4 marks

Recognize the sequence (as A.P or G.P) and find the general term of the sequence 4,2,1,\dfrac12,\dots

Solution

Test for an A.P.: the differences are not equal.
2-4=-2,\qquad 1-2=-1
Test for a G.P.: the ratios are equal.
\dfrac{a_2}{a_1}=\dfrac24=\dfrac12,\qquad\dfrac{a_3}{a_2}=\dfrac12,\qquad\dfrac{a_4}{a_3}=\dfrac{1/2}{1}=\dfrac12
So the sequence is a G.P. with a_1=4 and r=\dfrac12. Use a_n=a_1r^{n-1}.
a_n=4\left(\dfrac12\right)^{n-1}=2^2\cdot2^{1-n}=2^{3-n}
Check: n=4 gives 2^{-1}=\dfrac12.

a_n=4\left(\dfrac12\right)^{n-1}=2^{3-n}

2026 Annual I, Set B, Q.3 (b)4+2+2 = 2 marks

For the sequence 4,2,1,\dfrac12,\dots find the sum up to n terms.

Solution

The sequence is a G.P. with a_1=4 and r=\dfrac12 (part (a)). Since r<1 use S_n=\dfrac{a_1(1-r^n)}{1-r}.
S_n=\dfrac{4\left[1-\left(\frac12\right)^n\right]}{1-\frac12}
Divide by \dfrac12, that is multiply by 2.
S_n=8\left[1-\left(\dfrac12\right)^n\right]=8-2^{3-n}
Check: n=3 gives 8-1=7=4+2+1.

S_n=8\left[1-\left(\dfrac12\right)^n\right]

2026 Annual I, Set B, Q.3 (c)4+2+2 = 2 marks

For the sequence 4,2,1,\dfrac12,\dots find the sum up to infinity.

Solution

Here |r|=\dfrac12<1, so the infinite series has a sum.
S_\infty=\dfrac{a_1}{1-r}
Substitute.
S_\infty=\dfrac{4}{1-\frac12}=\dfrac{4}{\frac12}=8
This agrees with part (b): as n\to\infty, \left(\dfrac12\right)^n\to0 and S_n\to8.

S_\infty=8

2025 Annual I, Set B, Q.6 (a)8 for (a), (b) and (c) together = 8 marks

For an arithmetic-geometric series 1+5\left(\dfrac12\right)^1+9\left(\dfrac12\right)^2+13\left(\dfrac12\right)^3+\dots, find general term of the series.

Solution

The numbers 1,5,9,13,\dots are in A.P. The numbers 1,\dfrac12,\dfrac14,\dfrac18,\dots are in G.P.
a=1,\quad d=4,\quad b=1,\quad r=\dfrac12
Use T_n=[a+(n-1)d]\,b\,r^{n-1}.
T_n=[1+(n-1)4]\left(\dfrac12\right)^{n-1}
Simplify.
T_n=(4n-3)\left(\dfrac12\right)^{n-1}
Check: n=3 gives 9\left(\dfrac12\right)^2, the third term.

T_n=(4n-3)\left(\dfrac12\right)^{n-1}

2025 Annual I, Set B, Q.6 (b)8 for (a), (b) and (c) together = 8 marks

For an arithmetic-geometric series 1+5\left(\dfrac12\right)^1+9\left(\dfrac12\right)^2+13\left(\dfrac12\right)^3+\dots, sum the series up to n terms.

Solution

From part (a): a=1, d=4, r=\dfrac12. Use the book's formula for the sum of n terms.
S_n=\dfrac{a}{1-r}+\dfrac{dr(1-r^n)}{(1-r)^2}-\dfrac{(a+nd)r^n}{1-r}
Substitute. Note 1-r=\dfrac12 and (1-r)^2=\dfrac14.
S_n=\dfrac{1}{\frac12}+\dfrac{4\cdot\frac12\left(1-\frac1{2^n}\right)}{\frac14}-\dfrac{(1+4n)\frac1{2^n}}{\frac12}
Simplify each term.
S_n=2+8\left(1-\dfrac1{2^n}\right)-\dfrac{2(1+4n)}{2^n}
Collect the terms in \dfrac1{2^n}.
S_n=10-\dfrac{8+2+8n}{2^n}=10-\dfrac{8n+10}{2^n}
Check: n=1 gives 10-9=1; n=2 gives 10-\dfrac{26}{4}=3.5=1+2.5; n=3 gives 10-\dfrac{34}{8}=5.75=3.5+2.25.

S_n=10-\dfrac{8n+10}{2^n}

2025 Annual I, Set B, Q.6 (c)8 for (a), (b) and (c) together = 8 marks

For an arithmetic-geometric series 1+5\left(\dfrac12\right)^1+9\left(\dfrac12\right)^2+13\left(\dfrac12\right)^3+\dots, sum the series up to infinite terms.

Solution

Here |r|=\dfrac12<1, so the sum to infinity exists. Use the book's formula.
S_\infty=\dfrac{a}{1-r}+\dfrac{dr}{(1-r)^2}
Substitute a=1, d=4, r=\dfrac12.
S_\infty=\dfrac{1}{\frac12}+\dfrac{4\cdot\frac12}{\frac14}=2+8
Add.
S_\infty=10
This agrees with part (b): as n\to\infty, \dfrac{8n+10}{2^n}\to0, so S_n\to10.

S_\infty=10

2026 Annual I, Set D, MCQ 51 mark

5^{th} term of A.P for which (a_1=2,\ d=3) is:

A14Answer
B11
C13
D15

Why

a_5=a_1+(5-1)d=2+4(3)=14.

Option B, 11, is the 4th term (2+3\cdot3); it comes from using n-1 with n=4. Using 5d instead of 4d gives 17, which is not an option.

2025 Annual I, Set D, MCQ 51 mark

In an arithmetic series, if a=2, d=4 then S_{15}=

A240
B450Answer
C480
D510

Why

S_{15}=\dfrac{15}{2}\left[2(2)+14(4)\right]=\dfrac{15}{2}(60)=450.

Using 15d in place of (n-1)d=14d gives \dfrac{15}{2}(64)=480, the tempting option C.

2026 Annual I, Set D, MCQ 61 mark

If an A.P.=a_1,a_2,a_3,\dots then H.P.=

Aa_1,a_2,a_3,\dots
Ba_1+a_2+a_3+\dots
C\dfrac{1}{a_1+a_2+a_3+\dots}
D\dfrac{1}{a_1},\dfrac{1}{a_2},\dfrac{1}{a_3},\dotsAnswer

Why

A sequence is a harmonic sequence when the reciprocals of its terms form an A.P. So the H.P. made from the A.P. a_1,a_2,a_3,\dots is \dfrac1{a_1},\dfrac1{a_2},\dfrac1{a_3},\dots.

Options B and C are sums (series), not sequences, and C takes the reciprocal of the whole sum rather than of each term.

2025 Annual I, Set D, MCQ 61 mark

What is the sum of first two terms of an arithmetic-geometric sequence T_n=(2n+1)\cdot 3^{n-1} (n\in N)?

A5
B11
C18Answer
D23

Why

T_1=(3)(3^0)=3 and T_2=(5)(3^1)=15, so T_1+T_2=18.

Option A, 5, is only the arithmetic factor 2n+1 of the second term; the geometric factor 3^{n-1} must also be multiplied in.

2025 Annual I, Set D, MCQ 71 mark

For what value of x geometric series 1+2x+4x^2+8x^3+\dots converges? (r<1)

A\dfrac{1}{2}
B\dfrac{1}{4}Answer
C1
D\dfrac{2}{1}

Why

The common ratio is r=2x. The series converges when |r|<1, that is |2x|<1, so -\dfrac12<x<\dfrac12. Of the options only x=\dfrac14 lies in this interval (then r=\dfrac12).

At x=\dfrac12 the ratio is r=1, which is not less than 1, so the series does not converge (option A). Options C and D give r=2 and r=4.

2026 Annual I, Set D, Q.2 (iv) OR4 marks

Find the sum of 6 terms of sequence: 1+\dfrac12+\dfrac14+\dots

Solution

The ratio is constant, so this is a G.P.
a_1=1,\qquad r=\dfrac12
Use S_n=\dfrac{a_1(1-r^n)}{1-r} with n=6.
S_6=\dfrac{1-\left(\frac12\right)^6}{1-\frac12}=2\left(1-\dfrac1{64}\right)
Simplify.
S_6=2\cdot\dfrac{63}{64}=\dfrac{63}{32}
Check: 1+\dfrac12+\dfrac14+\dfrac18+\dfrac1{16}+\dfrac1{32}=\dfrac{32+16+8+4+2+1}{32}=\dfrac{63}{32}.

S_6=\dfrac{63}{32}

2025 Annual I, Set D, Q.2 (iv) OR (a)4 for (a) and (b) together = 4 marks

A harmonic sequence has 3rd term \dfrac{1}{8} and 5th term \dfrac{1}{20}. Find the general term.

Solution

The reciprocals are in A.P. with a_3=8 and a_5=20.
a_1+2d=8\quad\text{(i)},\qquad a_1+4d=20\quad\text{(ii)}
Subtract (i) from (ii).
2d=12\;\Rightarrow\;d=6
Put d=6 in (i).
a_1=8-12=-4
General term of the A.P.
a_n=-4+(n-1)6=6n-10
The general term of the H.P. is the reciprocal.
\dfrac{1}{a_n}=\dfrac{1}{6n-10}
Check: n=3 gives \dfrac18 and n=5 gives \dfrac1{20}.

\dfrac{1}{6n-10}

2025 Annual I, Set D, Q.2 (iv) OR (b)4 for (a) and (b) together = 4 marks

A harmonic sequence has 3rd term \dfrac{1}{8} and 5th term \dfrac{1}{20}. Find the 15th term of the sequence.

Solution

From part (a) the A.P. of reciprocals has a_n=6n-10. Find its 15th term.
a_{15}=6(15)-10=80
The 15th term of the H.P. is the reciprocal.
\dfrac{1}{80}

\dfrac{1}{80}

2026 Annual I, Set D, Q.2 (vi)4 marks

Find the 10^{th} term of an A.P with a=3 and the 7^{th} term of it is 15.

Solution

Use a_7=a+6d to find d.
3+6d=15\;\Rightarrow\;6d=12\;\Rightarrow\;d=2
Now a_{10}=a+9d.
a_{10}=3+9(2)=21
Check: the terms are 3,5,7,9,11,13,15 (7th term 15), then 17,19,21.

a_{10}=21

2025 Annual I, Set D, Q.2 (vi) (a)4 for (a) and (b) together = 4 marks

In an arithmetic sequence, first term is 12, and 10th term is 48. Find the common difference.

Solution

Use a_{10}=a_1+9d.
12+9d=48
Solve.
9d=36\;\Rightarrow\;d=4

d=4

2025 Annual I, Set D, Q.2 (vi) (b)4 for (a) and (b) together = 4 marks

In an arithmetic sequence, first term is 12, and 10th term is 48. Find the sum of the first 12 terms of the sequence.

Solution

From part (a), d=4. Use S_n=\dfrac n2\left[2a_1+(n-1)d\right] with n=12.
S_{12}=\dfrac{12}{2}\left[2(12)+11(4)\right]
Simplify.
S_{12}=6(24+44)=6(68)=408
Check: a_{12}=12+11(4)=56, and S_{12}=\dfrac{12}{2}(12+56)=408.

S_{12}=408

2026 Annual I, Set D, Q.2 (viii)4 marks

The 3^{rd} term of a geometric sequence is 16 and the 5^{th} term is 64. Find the common ratio and 1^{st} term. Also write the sequence for positive values.

Solution

Write the two terms with a_n=a_1r^{n-1}.
a_1r^2=16\quad\text{(i)},\qquad a_1r^4=64\quad\text{(ii)}
Divide (ii) by (i); a_1 cancels.
r^2=\dfrac{64}{16}=4\;\Rightarrow\;r=\pm2
Put r^2=4 in (i).
a_1(4)=16\;\Rightarrow\;a_1=4
For positive values take r=2. The sequence is
4,\ 8,\ 16,\ 32,\ 64,\ \dots\qquad a_n=4(2)^{n-1}=2^{n+1}
With r=-2 the sequence would be 4,-8,16,-32,64,\dots, which has the same 3rd and 5th terms but is not positive.

r=\pm2,\ a_1=4;\ \text{positive case: }4,8,16,32,64,\dots

2025 Annual I, Set D, Q.2 (viii)4 marks

A falcon dives from a height of 100 meters to catch its prey and then ascends back to 85% of its previous height after each unsuccessful attempt. Find the total vertical distance travelled by the falcon before stopping its attempts.

Solution

The first dive is 100 m. After each attempt the falcon climbs to 85% of the previous height, so the heights reached form a G.P.
h_1=100(0.85)=85,\qquad r=0.85,\qquad |r|<1
Take the attempts to go on without limit, so add all the heights with S_\infty=\dfrac{a_1}{1-r}.
85+72.25+\dots=\dfrac{85}{1-0.85}=\dfrac{85}{0.15}=\dfrac{1700}{3}
Each climb is followed by a dive of the same height, so these are counted twice; the first dive of 100 m is counted once.
\text{Distance}=100+2\left(\dfrac{1700}{3}\right)=\dfrac{300+3400}{3}=\dfrac{3700}{3}

\dfrac{3700}{3}\approx1233.3\,\mathrm{m}

2026 Annual I, Set D, Q.6 (a)4+4 = 4 marks

For the sequence 2,5,8,11,\dots find the general term. Also find the sum of first 20 terms.

Solution

The differences are all 3, so it is an A.P.
a_1=2,\qquad d=5-2=3
General term a_n=a_1+(n-1)d.
a_n=2+(n-1)3=3n-1
Sum of 20 terms with S_n=\dfrac n2\left[2a_1+(n-1)d\right].
S_{20}=\dfrac{20}{2}\left[4+19(3)\right]=10(61)
Simplify.
S_{20}=610
Check with a_{20}=3(20)-1=59: S_{20}=\dfrac{20}{2}(2+59)=610.

a_n=3n-1,\quad S_{20}=610

2025 Annual I, Set D, Q.6 (a)8 for (a), (b) and (c) together = 8 marks

For an arithmetic-geometric series 1(1)+3\left(\dfrac12\right)+5\left(\dfrac14\right)+7\left(\dfrac18\right)+9\left(\dfrac1{16}\right)+\dots, find n^{th} term of the series.

Solution

The numbers 1,3,5,7,9,\dots are in A.P. The numbers 1,\dfrac12,\dfrac14,\dfrac18,\dots are in G.P.
a=1,\quad d=2,\quad b=1,\quad r=\dfrac12
Use T_n=[a+(n-1)d]\,b\,r^{n-1}.
T_n=[1+(n-1)2]\left(\dfrac12\right)^{n-1}
Simplify.
T_n=(2n-1)\left(\dfrac12\right)^{n-1}
Check: n=4 gives 7\left(\dfrac12\right)^3=\dfrac78, the fourth term.

T_n=(2n-1)\left(\dfrac12\right)^{n-1}

2026 Annual I, Set D, Q.6 (b)4+4 = 4 marks

Find the sum of the n terms of the series 1.2+2.3+3.4+\dots

Solution

Read the pattern: each term is the product of a natural number and the next one. The kth term is
T_k=k(k+1)=k^2+k
Sum from k=1 to n and split with the properties of sigma.
S_n=\sum_{k=1}^{n}k^2+\sum_{k=1}^{n}k
Use \sum k^2=\dfrac{n(n+1)(2n+1)}{6} and \sum k=\dfrac{n(n+1)}{2}.
S_n=\dfrac{n(n+1)(2n+1)}{6}+\dfrac{n(n+1)}{2}
Take out the common factor \dfrac{n(n+1)}{6}.
S_n=\dfrac{n(n+1)}{6}\left[(2n+1)+3\right]=\dfrac{n(n+1)(2n+4)}{6}
Simplify.
S_n=\dfrac{n(n+1)(n+2)}{3}
Check: n=2 gives \dfrac{2\cdot3\cdot4}{3}=8=2+6.

S_n=\dfrac{n(n+1)(n+2)}{3}

2025 Annual I, Set D, Q.6 (b)8 for (a), (b) and (c) together = 8 marks

For the arithmetic-geometric series 1(1)+3\left(\dfrac12\right)+5\left(\dfrac14\right)+7\left(\dfrac18\right)+9\left(\dfrac1{16}\right)+\dots, sum the series up to n terms.

Solution

From part (a): a=1, d=2, r=\dfrac12. Use the book's formula for the sum of n terms.
S_n=\dfrac{a}{1-r}+\dfrac{dr(1-r^n)}{(1-r)^2}-\dfrac{(a+nd)r^n}{1-r}
Substitute. Note 1-r=\dfrac12 and (1-r)^2=\dfrac14.
S_n=\dfrac{1}{\frac12}+\dfrac{2\cdot\frac12\left(1-\frac1{2^n}\right)}{\frac14}-\dfrac{(1+2n)\frac1{2^n}}{\frac12}
Simplify each term.
S_n=2+4\left(1-\dfrac1{2^n}\right)-\dfrac{2(1+2n)}{2^n}
Collect the terms in \dfrac1{2^n}.
S_n=6-\dfrac{4+2+4n}{2^n}=6-\dfrac{4n+6}{2^n}
Check: n=1 gives 6-5=1; n=2 gives 6-\dfrac{14}{4}=2.5=1+1.5; n=3 gives 6-\dfrac{18}{8}=3.75=2.5+1.25.

S_n=6-\dfrac{4n+6}{2^n}

2025 Annual I, Set D, Q.6 (c)8 for (a), (b) and (c) together = 8 marks

For the arithmetic-geometric series 1(1)+3\left(\dfrac12\right)+5\left(\dfrac14\right)+7\left(\dfrac18\right)+9\left(\dfrac1{16}\right)+\dots, sum the series up to infinite terms.

Solution

Here |r|=\dfrac12<1, so the sum to infinity exists.
S_\infty=\dfrac{a}{1-r}+\dfrac{dr}{(1-r)^2}
Substitute a=1, d=2, r=\dfrac12.
S_\infty=\dfrac{1}{\frac12}+\dfrac{2\cdot\frac12}{\frac14}=2+4
Add.
S_\infty=6
This agrees with part (b): \dfrac{4n+6}{2^n}\to0 as n\to\infty.

S_\infty=6

Exam focus

Sum of an arithmetic series: S_n=\dfrac n2[2a_1+(n-1)d] for given a, d, n, and finding a_n and d from two given terms. (problem)
Asked in both years: 2026 Set B Q.2 (xi) and Set D Q.2 (vi) and Q.6 (a), 2025 Set B MCQ 5 and Q.2 (vi), Set D MCQ 5 and Q.2 (vi). Exercises 4.2 and 4.3 train the same two-equation method.
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Sum of n terms of a G.P. and the sum to infinity, for example 4+2+1+\dots or 3+9+27+\dots. (problem)
2026 Set B Q.2 (i), Q.2 (ix) OR and Q.3 (a) to (c), Set D Q.2 (iv) OR and Q.2 (viii); 2025 Set B MCQ 7 and Set D MCQ 7. Choose \dfrac{a(1-r^n)}{1-r} when r<1 and \dfrac{a(r^n-1)}{r-1} when r>1.
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Arithmetico-geometric series: find T_n, the sum of n terms and the sum to infinity. (long)
8-mark Q.6 (a), (b), (c) in both 2025 sets (series with r=\dfrac12), plus 2025 Set D MCQ 6. Not asked in 2026, but the formulas are quick once a, d, r are identified.
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Harmonic sequence: find the general term or a term from two given terms, by taking reciprocals to an A.P. (problem)
2025 Set B Q.2 (iv) OR and Set D Q.2 (iv) OR (a), (b); 2025 Set B MCQ 6; 2026 Set B MCQ 11 and Set D MCQ 6. Exercise 4.6 has the same questions.
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Total distance travelled by a bouncing ball (infinite G.P. application): first fall plus twice the rebound sum. (problem)
2025 Set B Q.2 (viii) (ball, 80%) and Set D Q.2 (viii) (falcon, 85%); Exercise 4.5 Q.15 is the same type.
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Sum a series using sigma formulas, for example 1.2+2.3+3.4+\dots. (problem)
2026 Set D Q.6 (b). Find the kth term, then use \sum k and \sum k^2. Exercise 4.7 Q.14 to Q.22 practise this.
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Identify a sequence as A.P. or G.P. and write its general term. (problem)
2026 Set B Q.3 (a) and Set D Q.2 (viii); test for equal differences or equal ratios first.
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Standard formulas as MCQs: a_n=a+(n-1)d, 1+3+5+\dots+(2n-1)=n^2, infinite sum \dfrac{a}{1-r}. (mcq)
2026 Set B MCQ 10, Set D MCQ 5; 2025 Set B MCQ 5 and MCQ 8. One mark each, so learn the formula sheet.
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