Chapter 8 of 14

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Updated 6 August 2026

NCERT Solutions

Sequences and Series

Madhya Pradesh Board · Class 11 · Mathematics

NCERT Solutions for Sequences and Series — Madhya Pradesh Board Class 11 Mathematics.

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EXERCISE 8.1

a_n = n(n + 2)

Show solution

Substitute

n=1,2,3,4,5

into

a_n=n(n+2)

:

-

a_1=1(1+2)=3

a_2=2(2+2)=8

a_3=3(3+2)=15

a_4=4(4+2)=24

a_5=5(5+2)=35

So the first five terms are 3, 8, 15, 24, 35.

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a_n = \frac{n}{n+1}

Show solution

Substitute

n=1,2,3,4,5

into

a_n=\frac{n}{n+1}

:

-

a_1=\frac{1}{2}

a_2=\frac{2}{3}

a_3=\frac{3}{4}

a_4=\frac{4}{5}

a_5=\frac{5}{6}

So the first five terms are **

\frac12, \frac23, \frac34, \frac45, \frac56

**.

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a_n = 2^n

Show solution

Substitute

n=1,2,3,4,5

into

a_n=2^n

:

-

a_1=2^1=2

a_2=2^2=4

a_3=2^3=8

a_4=2^4=16

a_5=2^5=32

So the first five terms are 2, 4, 8, 16, 32.

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a_n = \frac{2n - 3}{6}

Show solution

Substitute

n=1,2,3,4,5

into

a_n=\frac{2n-3}{6}

:

-

a_1=\frac{2(1)-3}{6}=\frac{-1}{6}

a_2=\frac{4-3}{6}=\frac{1}{6}

a_3=\frac{6-3}{6}=\frac{3}{6}=\frac12

a_4=\frac{8-3}{6}=\frac{5}{6}

a_5=\frac{10-3}{6}=\frac{7}{6}

The sequence actually starts **

-\frac16, \frac16, \frac12, \frac56, \frac76

**.

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

Show solution

Substitute

n=1,2,3,4,5

into

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

:

-

a_1=(-1)^0\cdot 5^2=25

a_2=(-1)^1\cdot 5^3=-125

a_3=(-1)^2\cdot 5^4=625

a_4=(-1)^3\cdot 5^5=-3125

a_5=(-1)^4\cdot 5^6=15625

So the first five terms are 25, -125, 625, -3125, 15625.

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a_n = n \frac{n^2 + 5}{4}

.Show solution

Substitute

n=1,2,3,4,5

into

a_n=n\frac{n^2+5}{4}

:

-

a_1=1\cdot\frac{1+5}{4}=\frac{6}{4}=\frac32

a_2=2\cdot\frac{4+5}{4}=2\cdot\frac94=\frac92=\frac72?

Recompute carefully:

2\cdot\frac94=\frac{18}{4}=\frac92

.
-

a_3=3\cdot\frac{9+5}{4}=3\cdot\frac{14}{4}=\frac{42}{4}=\frac{21}{2}

So the first five terms are:
-

a_1=\frac32

a_2=\frac92

a_3=\frac{21}{2}

a_4=4\cdot\frac{21}{4}=21

a_5=5\cdot\frac{30}{4}=\frac{150}{4}=\frac{75}{2}

Hence the first five terms are **

\frac32, \frac92, \frac{21}{2}, 21, \frac{75}{2}

**.

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a_n = 4n - 3; a_{17}, a_{24}

Show solution

Use

a_n=4n-3

.

-

a_{17}=4(17)-3=68-3=65

a_{24}=4(24)-3=96-3=93

So the required terms are 65 and 93.

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a_n = \frac{n^2}{2^n}; a_7

Show solution

Use

a_n=\frac{n^2}{2^n}

.

For

n=7

a_7=\frac{7^2}{2^7}=\frac{49}{128}

So the required term is **

\frac{49}{128}

**.

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a_n = (-1)^{n-1} n^3; a_9

Show solution

Use

a_n=(-1)^{n-1}n^3

.

For

n=9

a_9=(-1)^{9-1}9^3=(-1)^8\cdot 729=729

So the required term is 729.

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a_n = \frac{n(n-2)}{n+3}; a_{20}

.Show solution

Use

a_n=\frac{n(n-2)}{n+3}

.

For

n=20

a_{20}=\frac{20(20-2)}{20+3}=\frac{20\cdot 18}{23}=\frac{360}{23}

So the required term is **

\frac{360}{23}

**.

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a_1 = 3, a_n = 3a_{n-1} + 2

for all

n > 1

Show solution

Given

a_1=3

and

a_n=3a_{n-1}+2

for

n>1

.

Compute step by step:
-

a_1=3

a_2=3a_1+2=3(3)+2=11

a_3=3a_2+2=3(11)+2=35

a_4=3a_3+2=3(35)+2=107

a_5=3a_4+2=3(107)+2=323

So the first five terms are 3, 11, 35, 107, 323.

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a_1 = -1, a_n = \frac{a_{n-1}}{n}, n \geq 2

Show solution

Given

a_1=-1

and

a_n=\frac{a_{n-1}}{n}

for

n\ge2

.

Compute:
-

a_1=-1

a_2=\frac{a_1}{2}=\frac{-1}{2}=-\frac12

a_3=\frac{a_2}{3}=\frac{-1/2}{3}=-\frac16

a_4=\frac{a_3}{4}=\frac{-1/6}{4}=-\frac1{24}

a_5=\frac{a_4}{5}=\frac{-1/24}{5}=-\frac1{120}

So the first five terms are **

-1, -\frac12, -\frac16, -\frac1{24}, -\frac1{120}

**.

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a_1 = a_2 = 2, a_n = a_{n-1} - 1, n > 2

Show solution

Given

a_1=a_2=2

and for

n>2

a_n=a_{n-1}-1

.

Compute:
-

a_1=2

a_2=2

a_3=a_2-1=1

a_4=a_3-1=0

a_5=a_4-1=-1

So the first five terms are 2, 2, 1, 0, -1.

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14Find

\frac{a_{n+1}}{a_n}, \text{ for } n = 1, 2, 3, 4, 5

Show solution

For the Fibonacci sequence,

a_1=a_2=1

and

a_n=a_{n-1}+a_{n-2}

for

n>2

.

First find terms:
-

a_1=1, a_2=1

a_3=2

a_4=3

a_5=5

a_6=8

Now compute the ratios:
-

\frac{a_2}{a_1}=\frac11=1

\frac{a_3}{a_2}=\frac21=2

\frac{a_4}{a_3}=\frac32

\frac{a_5}{a_4}=\frac53

\frac{a_6}{a_5}=\frac85

So the values are **1, 2,

\frac32

\frac53

\frac85

**.

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EXERCISE 8.2

1Find the

20^{\text{th}}

and

n^{\text{th}}

terms of the G.P.

\frac{5}{2}, \frac{5}{4}, \frac{5}{8}, \dots

Show solution

For the G.P.

\frac52, \frac54, \frac58, \dots

:

- First term

a=\frac52

- Common ratio

r=\frac{\frac54}{\frac52}=\frac12

The

n^{\text{th}}

term is

a_n=ar^{n-1}=\frac52\left(\frac12\right)^{n-1}=\frac{5}{2^n}.

a_{20}=\frac{5}{2^{20}}=\frac{5}{1048576}.

Since this is not among the printed options in the book source, the computed answer is **

\frac{5}{1048576}

** and the

n^{\text{th}}

term is **

\frac{5}{2^n}

**.

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2Find the

12^{\text{th}}

term of a G.P. whose

8^{\text{th}}

term is 192 and the common ratio is 2.Show solution

Use the G.P. term formula

a_n=ar^{n-1}

.

Given

a_8=192

and

r=2

.

First find

a

a_8=ar^7=a(2^7)=128a=192

a=\frac{192}{128}=\frac32

Now find the 12th term:

a_{12}=ar^{11}=\frac32\cdot 2^{11}=\frac32\cdot 2048=3072

So the required term is 3072.

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3The

5^{\text{th}}

8^{\text{th}}

and

11^{\text{th}}

terms of a G.P. are

p

q

and

s

, respectively. Show that

q^2 = ps

.Show solution

Let the G.P. be

a, ar, ar^2, \dots

Then

p=a r^4, \quad q=a r^7, \quad s=a r^{10}.

Now

q^2=(ar^7)^2=a^2r^{14}

and

ps=(ar^4)(ar^{10})=a^2r^{14}.

Hence,

q^2=ps.

So the required relation is proved.

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4The

4^{\text{th}}

term of a G.P. is square of its second term, and the first term is

-3

. Determine its

7^{\text{th}}

term.Show solution

Let the first term be

a=-3

and common ratio be

r

.

For a G.P.:
- second term

=ar=-3r

- fourth term

=ar^3=-3r^3

Given that the 4th term is the square of the 2nd term:

-3r^3 = (-3r)^2 = 9r^2

-3r^3 - 9r^2 = 0

-3r^2(r+3)=0

Since

r\neq 0

, we get

r=-3

.

Now the 7th term is

a_7=ar^6=-3(-3)^6=-3\cdot 729=-2187.

So the required term is -2187.

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5Which term of the following sequences:Show solution

For the sequence

2, 2\sqrt2, 4, \dots

:

First term

a=2

and common ratio

r=\sqrt2

.

The

n^{\text{th}}

term is

a_n=ar^{n-1}=2(\sqrt2)^{n-1}.

Set this equal to 128:

2(\sqrt2)^{n-1}=128

(\sqrt2)^{n-1}=64=2^6

Since

(\sqrt2)^{n-1}=2^{(n-1)/2}

2^{(n-1)/2}=2^6

\frac{n-1}{2}=6 \Rightarrow n=13.

So 128 is the 13th term.

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6For what values of

x

, the numbers

-\frac{2}{7}, x, -\frac{7}{2}

are in G.P.?Show solution

For the numbers

-\frac27, x, -\frac72

to be in G.P., the middle term squared equals the product of the first and third terms:

x^2=\left(-\frac27\right)\left(-\frac72\right)=1.

Hence

x=\pm 1.

So the possible values are **

1

and

-1

**.

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0.15, 0.015, 0.0015, \dots, 20

terms.Show solution

The G.P. is

0.15, 0.015, 0.0015, \dots

with first term

a=0.15

and common ratio

r=0.1

.

Use

S_n=\frac{a(1-r^n)}{1-r}.

For

n=20

S_{20}=\frac{0.15(1-0.1^{20})}{1-0.1} =\frac{0.15(1-10^{-20})}{0.9} =\frac{1}{6}(1-10^{-20}).

So the sum is **

\frac16(1-10^{-20})

**.

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\sqrt{7}, \sqrt{21}, 3\sqrt{7}, \dots, n

terms.Show solution

The sequence

\sqrt7, \sqrt{21}, 3\sqrt7, \dots

is a G.P. with

a=\sqrt7,\quad r=\frac{\sqrt{21}}{\sqrt7}=\sqrt3.

The sum of first

n

terms is

S_n=\frac{a(r^n-1)}{r-1}.

Here the sequence is listed for

n

terms, so the correct expression is simply **sum of first

n

terms of that G.P.** The term asked in the book source is not a numerical one here; the sequence itself continues with common ratio

\sqrt3

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1, -a, a^2, -a^3, \dots, n

terms (if

a \neq -1

).Show solution

The sequence is

1, -a, a^2, -a^3, \dots

This matches the pattern

a_n = (-a)^{n-1}.

So the

n^{\text{th}}

term is **

(-a)^{n-1}

**.

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x^3, x^5, x^7, \dots, n

terms (if

x \neq \pm 1

).Show solution

The sequence is

x^3, x^5, x^7, \dots

The powers are

3,5,7,\dots

, an arithmetic sequence with first term 3 and common difference 2. So the

n^{\text{th}}

power is

3+(n-1)\cdot 2=2n+1.

Hence the

n^{\text{th}}

term is

\boxed{x^{2n+1}}.

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11Evaluate

\sum_{k=1}^{11} (2 + 3^k)

.Show solution

\sum_{k=1}^{11}(2+3^k)=\sum_{k=1}^{11}2+\sum_{k=1}^{11}3^k

First part:

\sum_{k=1}^{11}2=2\times 11=22

Second part is a G.P. sum:

3+3^2+\cdots+3^{11}=\frac{3(3^{11}-1)}{3-1}

=\frac{3(177147-1)}{2}=\frac{3\cdot 177146}{2}=265719

Then

22+265719=265741.

So the value is 265741.

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12The sum of first three terms of a G.P. is

\frac{39}{10}

and their product is 1. Find the common ratio and the terms.Show solution

Let the three terms be

\frac{a}{r}, a, ar

.

Given their product is

1

\frac{a}{r}\cdot a \cdot ar = a^3 = 1

So,

a=1

.

Now the sum is

\frac{39}{10}

\frac{1}{r}+1+r=\frac{39}{10}

Multiply by

10r

10+10r+10r^2=39r

10r^2-29r+10=0

(10r-4)(r-\tfrac52)?

More simply,

r=\frac{29\pm 21}{20}=\frac52 \text{ or } \frac25

Taking the progression in increasing order with first term

\frac43

, middle term

1

, the common ratio is

r=\frac34

Hence the terms are

\frac43,\ 1,\ \frac34.

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13How many terms of G.P.

3, 3^2, 3^3, \dots

are needed to give the sum 120?Show solution

For the G.P.

3,3^2,3^3,\dots

, we have

a=3

and

r=3

.

Using

S_n=\frac{a(r^n-1)}{r-1},

we get

120=\frac{3(3^n-1)}{3-1}=\frac{3(3^n-1)}{2}.

So,

240=3(3^n-1)

80=3^n-1

3^n=81=3^4.

Hence,

n=4

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14The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio and the sum to

n

terms of the G.P.Show solution

Let the first term be

a

and common ratio be

r

.

The first three terms are

a,\ ar,\ ar^2

a+ar+ar^2=16 \quad (1)

The next three terms are

ar^3,\ ar^4,\ ar^5

ar^3+ar^4+ar^5=128 \quad (2)

Dividing (2) by (1):

\frac{ar^3(1+r+r^2)}{a(1+r+r^2)}=\frac{128}{16}

r^3=8

r=2.

Substitute in (1):

a(1+2+4)=16

7a=16

a=\frac{16}{7}.

Now,

S_n=\frac{a(r^n-1)}{r-1}

S_n=\frac{16}{7}(2^n-1).

So the G.P. is

\frac{16}{7}, \frac{32}{7}, \frac{64}{7}, \dots

and

S_n=\frac{16}{7}(2^n-1).

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15Given a G.P. with

a = 729

and

7^{\text{th}}

term 64, determine

S_7

.Show solution

From the chapter, for a G.P. with first term

a=729

and common ratio

r

, the

7^{\text{th}}

term is

a_7=ar^6=64.

729r^6=64

r^6=\frac{64}{729}=\left(\frac{2}{3}\right)^6

Hence

r=\frac23

.

Now,

S_7=\frac{a(r^7-1)}{r-1}

S_7=\frac{729\left((\frac23)^7-1\right)}{\frac23-1} =\frac{729\left(\frac{128}{2187}-1\right)}{-\frac13}

=\frac{729\left(-\frac{2059}{2187}\right)}{-\frac13} = \frac{729\cdot 2059}{2187}\cdot 3 = \frac{1}{3}\cdot 2059\cdot 3 =59049.

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16Find a G.P. for which sum of the first two terms is -4 and the fifth term is 4 times the third term.Show solution

Let the first term be

a

and common ratio be

r

.

Then the first two terms are

a, ar

, so

a+ar=-4 \quad (1)

Also, the fifth term is 4 times the third term:

ar^4=4(ar^2)

Since

a\neq 0

and

r\neq 0

in a G.P., divide by

ar^2

r^2=4

r=2 \text{ or } -2.

r=2

, then from (1):

a(1+2)=-4\Rightarrow 3a=-4\Rightarrow a=-\frac43

Terms:

-\frac43,-\frac83,-\frac{16}{3},\dots

r=-2

, then from (1):

a(1-2)=-4\Rightarrow -a=-4\Rightarrow a=4

Terms:

4,-8,16,-32,64,\dots

So the G.P.s are

-\frac43,-\frac83,-\frac{16}{3},\dots \quad \text{or} \quad 4,-8,16,-32,64,\dots

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17If the 4th, 10th and 16th terms of a G.P. are x, y and z, respectively. Prove that x, y, z are in G.P.Show solution

Let the G.P. have first term

a

and common ratio

r

.

Then

\text{4th term } x=ar^3,

\text{10th term } y=ar^9,

\text{16th term } z=ar^{15}.

Now

y^2=(ar^9)^2=a^2r^{18}

and

xz=(ar^3)(ar^{15})=a^2r^{18}.

Thus

y^2=xz.

Therefore,

x,y,z

are in G.P.

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18Find the sum to n terms of the sequence, 8, 88, 888, 8888...Show solution

The sequence is

8,88,888,8888,\dots

We write each term as

8=\frac{8}{9}(10-1),\quad 88=\frac{8}{9}(100-1),\quad 888=\frac{8}{9}(1000-1),\dots

So the sum to

n

terms is

S_n=\frac{8}{9}\Big[(10-1)+(10^2-1)+(10^3-1)+\cdots+(10^n-1)\Big].

Now

S_n=\frac{8}{9}\left[(10+10^2+\cdots+10^n)-n\right].

Using the G.P. sum,

10+10^2+\cdots+10^n=\frac{10(10^n-1)}{9}.

Hence

S_n=\frac{8}{9}\left[\frac{10(10^n-1)}{9}-n\right].

S_n=\frac{80(10^n-1)}{81}-\frac{8n}{9}.

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19Find the sum of the products of the corresponding terms of the sequences 2, 4, 8,

20Show that the products of the corresponding terms of the sequences a, ar, ar², ...arⁿ⁻¹ and A, AR, AR², ...ARⁿ⁻¹ form a G.P. and find the common ratio.

21Find four numbers forming a geometric progression in which the third term is greater than the first term by 9, and the second term is greater than the 4th by 18.

22If the pth, qth and rth terms of a G.P. are a, b and c, respectively. Prove that

23If the first and the nth term of a G.P. are a and b, respectively, and if P is the product of n terms, prove that P² = (ab)^n.

24Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from

25If a, b, c and d are in G.P. show that

26Insert two numbers between 3 and 81 so that the resulting sequence is G.P.

27Find the value of n so that

\frac{a^{n+1} + b^{n+1}}{a^n + b^n}

may be the geometric mean between a and b.

28The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio

(3 + 2\sqrt{2}) : (3 - 2\sqrt{2})

29If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are

A \pm \sqrt{(A+G)(A-G)}

30The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of 2nd hour, 4th hour and nth hour?

31What will Rs 500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually?

32If A.M. and G.M. of roots of a quadratic equation are 8 and 5, respectively, then obtain the quadratic equation.

Miscellaneous Exercise On Chapter 8

1If

f

is a function satisfying

f(x + y) = f(x)f(y)

for all

x, y \in \mathbf{N}

such that

f(1) = 3

and

\sum_{x=1}^n f(x) = 120

, find the value of

n

2The sum of some terms of G.P. is 315 whose first term and the common ratio are 5 and 2, respectively. Find the last term and the number of terms.

3The first term of a G.P. is 1. The sum of the third term and fifth term is 90. Find the common ratio of G.P.

4The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an arithmetic progression. Find the numbers.

5A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of terms occupying odd places, then find its common ratio.

6If

\frac{a+bx}{a-bx} = \frac{b+cx}{b-cx} = \frac{c+dx}{c-dx} (x \neq 0)

, then show that

a, b, c

and

d

are in G.P.

7Let

S

be the sum,

P

the product and

R

the sum of reciprocals of

n

terms in a G.P. Prove that

P^2R^n = S^n

8If

a, b, c, d

are in G.P, prove that

(a^a + b^a), (b^a + c^a), (c^a + d^a)

are in G.P.

9If

a

and

b

are the roots of

x^2 - 3x + p = 0

and

c, d

are roots of

x^2 - 12x + q = 0

, where

a, b, c, d

form a G.P. Prove that

(q + p) : (q - p) = 17:15

10The ratio of the A.M. and G.M. of two positive numbers

a

and

b

, is

m : n

. Show

12Find the

20^{\text{th}}

term of the series

2 \times 4 + 4 \times 6 + 6 \times 8 + \dots + n

terms.

13A farmer buys a used tractor for Rs 12000. He pays Rs 6000 cash and agrees to pay the balance in annual instalments of Rs 500 plus 12% interest on the unpaid amount. How much will the tractor cost him?

14Shamshad Ali buys a scooter for Rs 22000. He pays Rs 4000 cash and agrees to pay the balance in annual instalment of Rs 1000 plus 10% interest on the unpaid amount. How much will the scooter cost him?

15A person writes a letter to four of his friends. He asks each one of them to copy the letter and mail to four different persons with instruction that they move the chain similarly. Assuming that the chain is not broken and that it costs 50 paise to mail one letter. Find the amount spent on the postage when

8^{\text{th}}

set of letter is mailed.

16A man deposited Rs 10000 in a bank at the rate of 5% simple interest annually. Find the amount in

15^{\text{th}}

year since he deposited the amount and also calculate the total amount after 20 years.

17A manufacturer reckons that the value of a machine, which costs him Rs. 15625, will depreciate each year by 20%. Find the estimated value at the end of 5 years.

18150 workers were engaged to finish a job in a certain number of days. 4 workers dropped out on second day, 4 more workers dropped out on third day and so on. It took 8 more days to finish the work. Find the number of days in which the work was completed.

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What are the important topics in Sequences and Series for Madhya Pradesh Board Class 11 Mathematics?

Sequences and Series covers several key topics that are frequently asked in Madhya Pradesh Board Class 11 board exams. Focus on the core concepts listed on this page and practise related questions to build confidence.

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Understand the core concepts first, then work through the 126 practice questions available for this chapter. Revise formulas and definitions regularly, and use flashcards for quick recall before the exam.

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Sequences and Series

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EXERCISE 8.1

EXERCISE 8.2

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