Chapter 8 of 14

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Updated 11 June 2026

NCERT Solutions

Sequences and Series

Tripura Board · Class 11 · Mathematics

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

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64 Questions Solved · 3 Sections

Exercise 8.1

1Write the first five terms of the sequence whose

n^{\text{th}}

term is

a_n = n(n+2)

.Show solution

Given:

a_n = n(n+2)

Substituting $n = 1, 2, 3, 4, 5$ :

a_1 = 1(1+2) = 1 \times 3 = 3

a_2 = 2(2+2) = 2 \times 4 = 8

a_3 = 3(3+2) = 3 \times 5 = 15

a_4 = 4(4+2) = 4 \times 6 = 24

a_5 = 5(5+2) = 5 \times 7 = 35

The first five terms are: $3, 8, 15, 24, 35$ .

2Write the first five terms of the sequence whose

n^{\text{th}}

term is

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

.Show solution

Given:

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

Substituting $n = 1, 2, 3, 4, 5$ :

a_1 = \frac{1}{1+1} = \frac{1}{2}

a_2 = \frac{2}{2+1} = \frac{2}{3}

a_3 = \frac{3}{3+1} = \frac{3}{4}

a_4 = \frac{4}{4+1} = \frac{4}{5}

a_5 = \frac{5}{5+1} = \frac{5}{6}

The first five terms are: $\dfrac{1}{2},\ \dfrac{2}{3},\ \dfrac{3}{4},\ \dfrac{4}{5},\ \dfrac{5}{6}$ .

3Write the first five terms of the sequence whose

n^{\text{th}}

term is

a_n = 2^n

.Show solution

Given:

a_n = 2^n

Substituting $n = 1, 2, 3, 4, 5$ :

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

The first five terms are: $2, 4, 8, 16, 32$ .

4Write the first five terms of the sequence whose

n^{\text{th}}

term is

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

.Show solution

Given:

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

Substituting $n = 1, 2, 3, 4, 5$ :

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

a_2 = \frac{2(2)-3}{6} = \frac{1}{6}

a_3 = \frac{2(3)-3}{6} = \frac{3}{6} = \frac{1}{2}

a_4 = \frac{2(4)-3}{6} = \frac{5}{6}

a_5 = \frac{2(5)-3}{6} = \frac{7}{6}

The first five terms are: $-\dfrac{1}{6},\ \dfrac{1}{6},\ \dfrac{1}{2},\ \dfrac{5}{6},\ \dfrac{7}{6}$ .

5Write the first five terms of the sequence whose

n^{\text{th}}

term is

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

.Show solution

Given:

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

Substituting $n = 1, 2, 3, 4, 5$ :

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

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

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

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

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

The first five terms are: $25,\ -125,\ 625,\ -3125,\ 15625$ .

6Write the first five terms of the sequence whose

n^{\text{th}}

term is

a_n = n \cdot \dfrac{n^2+5}{4}

.Show solution

Given:

a_n = n \cdot \dfrac{n^2+5}{4}

Substituting $n = 1, 2, 3, 4, 5$ :

a_1 = 1 \cdot \frac{1+5}{4} = \frac{6}{4} = \frac{3}{2}

a_2 = 2 \cdot \frac{4+5}{4} = 2 \cdot \frac{9}{4} = \frac{9}{2}

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

a_4 = 4 \cdot \frac{16+5}{4} = 4 \cdot \frac{21}{4} = 21

a_5 = 5 \cdot \frac{25+5}{4} = 5 \cdot \frac{30}{4} = \frac{75}{2}

The first five terms are: $\dfrac{3}{2},\ \dfrac{9}{2},\ \dfrac{21}{2},\ 21,\ \dfrac{75}{2}$ .

7Find

a_{17}

and

a_{24}

for the sequence whose

n^{\text{th}}

term is

a_n = 4n - 3

.Show solution

Given:

a_n = 4n - 3

Finding $a_{17}$ : Put

n = 17

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

Finding $a_{24}$ : Put

n = 24

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

Therefore, $a_{17} = 65$ and $a_{24} = 93$ .

8Find

a_7

for the sequence whose

n^{\text{th}}

term is

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

.Show solution

Given:

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

Finding $a_7$ : Put

n = 7

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

Therefore, $a_7 = \dfrac{49}{128}$ .

9Find

a_9

for the sequence whose

n^{\text{th}}

term is

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

.Show solution

Given:

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

Finding $a_9$ : Put

n = 9

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

Therefore, $a_9 = 729$ .

10Find

a_{20}

for the sequence whose

n^{\text{th}}

term is

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

.Show solution

Given:

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

Finding $a_{20}$ : Put

n = 20

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

Therefore, $a_{20} = \dfrac{360}{23}$ .

11Write the first five terms of the sequence defined by

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

for all

n &gt; 1

, and obtain the corresponding series.Show solution

Given:

a_1 = 3

and

a_n = 3a_{n-1} + 2

for

n &gt; 1

.

Finding the terms:

a_1 = 3

a_2 = 3a_1 + 2 = 3(3) + 2 = 9 + 2 = 11

a_3 = 3a_2 + 2 = 3(11) + 2 = 33 + 2 = 35

a_4 = 3a_3 + 2 = 3(35) + 2 = 105 + 2 = 107

a_5 = 3a_4 + 2 = 3(107) + 2 = 321 + 2 = 323

The first five terms are: $3, 11, 35, 107, 323$ .

Corresponding series:

3 + 11 + 35 + 107 + 323 + \ldots

12Write the first five terms of the sequence defined by

a_1 = -1,\ a_n = \dfrac{a_{n-1}}{n}

for

n \geq 2

, and obtain the corresponding series.Show solution

Given:

a_1 = -1

and

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

for

n \geq 2

.

Finding the terms:

a_1 = -1

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

a_3 = \frac{a_2}{3} = \frac{-1/2}{3} = \frac{-1}{6}

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

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

The first five terms are: $-1,\ -\dfrac{1}{2},\ -\dfrac{1}{6},\ -\dfrac{1}{24},\ -\dfrac{1}{120}$ .

Corresponding series:

-1 + \left(-\dfrac{1}{2}\right) + \left(-\dfrac{1}{6}\right) + \left(-\dfrac{1}{24}\right) + \left(-\dfrac{1}{120}\right) + \ldots

13Write the first five terms of the sequence defined by

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

for

n &gt; 2

, and obtain the corresponding series.Show solution

Given:

a_1 = a_2 = 2

and

a_n = a_{n-1} - 1

for

n &gt; 2

.

Finding the terms:

a_1 = 2

a_2 = 2

a_3 = a_2 - 1 = 2 - 1 = 1

a_4 = a_3 - 1 = 1 - 1 = 0

a_5 = a_4 - 1 = 0 - 1 = -1

The first five terms are: $2, 2, 1, 0, -1$ .

Corresponding series:

2 + 2 + 1 + 0 + (-1) + \ldots

14The Fibonacci sequence is defined by

a_1 = a_2 = 1

and

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

for

n &gt; 2

. Find

\dfrac{a_{n+1}}{a_n}

for

n = 1, 2, 3, 4, 5

.Show solution

Given:

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

for

n &gt; 2

.

Finding the Fibonacci terms:

a_1 = 1,\quad a_2 = 1

a_3 = a_2 + a_1 = 1 + 1 = 2

a_4 = a_3 + a_2 = 2 + 1 = 3

a_5 = a_4 + a_3 = 3 + 2 = 5

a_6 = a_5 + a_4 = 5 + 3 = 8

Now computing $\dfrac{a_{n+1}}{a_n}$ :

For

n = 1

\dfrac{a_2}{a_1} = \dfrac{1}{1} = 1

For

n = 2

\dfrac{a_3}{a_2} = \dfrac{2}{1} = 2

For

n = 3

\dfrac{a_4}{a_3} = \dfrac{3}{2}

For

n = 4

\dfrac{a_5}{a_4} = \dfrac{5}{3}

For

n = 5

\dfrac{a_6}{a_5} = \dfrac{8}{5}

Therefore, the values of $\dfrac{a_{n+1}}{a_n}$ for $n = 1, 2, 3, 4, 5$ are $1,\ 2,\ \dfrac{3}{2},\ \dfrac{5}{3},\ \dfrac{8}{5}$ respectively.

Exercise 8.2

1Find the

20^{\text{th}}

and

n^{\text{th}}

terms of the G.P.

\dfrac{5}{2}, \dfrac{5}{4}, \dfrac{5}{8}, \ldots

Show solution

Given G.P.:

\dfrac{5}{2}, \dfrac{5}{4}, \dfrac{5}{8}, \ldots

First term:

a = \dfrac{5}{2}

Common ratio:

r = \dfrac{5/4}{5/2} = \dfrac{5}{4} \times \dfrac{2}{5} = \dfrac{1}{2}

Formula for $n^{\text{th}}$ term:

a_n = ar^{n-1}

a_n = \frac{5}{2} \cdot \left(\frac{1}{2}\right)^{n-1} = \frac{5}{2} \cdot \frac{1}{2^{n-1}} = \frac{5}{2^n}

For the $20^{\text{th}}$ term: Put

n = 20

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

Therefore, $a_n = \dfrac{5}{2^n}$ and $a_{20} = \dfrac{5}{2^{20}}$ .

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

Given:

a_8 = 192

r = 2

Using $a_n = ar^{n-1}$ :

a_8 = a \cdot r^7 = a \cdot 2^7 = 128a = 192

a = \frac{192}{128} = \frac{3}{2}

Finding $a_{12}$ :

a_{12} = a \cdot r^{11} = \frac{3}{2} \cdot 2^{11} = \frac{3}{2} \cdot 2048 = 3 \times 1024 = 3072

Therefore, the $12^{\text{th}}$ term is $3072$ .

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

Given: In a G.P. with first term

a

and common ratio

r

a_5 = p \Rightarrow ar^4 = p

a_8 = q \Rightarrow ar^7 = q

a_{11} = s \Rightarrow ar^{10} = s

Now compute $ps$ :

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

Compute $q^2$ :

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

Therefore:

q^2 = ps \qquad \textbf{(Proved)}

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

Given: First term

a = -3

, and

a_4 = (a_2)^2

.

Using $a_n = ar^{n-1}$ :

a_4 = ar^3 = -3r^3

a_2 = ar = -3r

Condition:

a_4 = (a_2)^2

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

-3r^3 = 9r^2

-3r = 9 \quad (\text{dividing both sides by } r^2, r \neq 0)

r = -3

Finding $a_7$ :

a_7 = ar^6 = (-3)(-3)^6 = (-3)^7 = -2187

Therefore, the $7^{\text{th}}$ term is $-2187$ .

5Which term of the following sequences: (a)

2, 2\sqrt{2}, 4, \ldots

is 128? (b)

\sqrt{3}, 3, 3\sqrt{3}, \ldots

is 729? (c)

\dfrac{1}{3}, \dfrac{1}{9}, \dfrac{1}{27}, \ldots

\dfrac{1}{19683}

?Show solution

(a)

2, 2\sqrt{2}, 4, \ldots

First term

a = 2

, common ratio

r = \dfrac{2\sqrt{2}}{2} = \sqrt{2}

.

Let

a_n = 128

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

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

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

\frac{n-1}{2} = 6 \Rightarrow n - 1 = 12 \Rightarrow n = 13

128 is the $13^{\text{th}}$ term.

---

(b)

\sqrt{3}, 3, 3\sqrt{3}, \ldots

First term

a = \sqrt{3}

, common ratio

r = \dfrac{3}{\sqrt{3}} = \sqrt{3}

.

Let

a_n = 729

\sqrt{3} \cdot (\sqrt{3})^{n-1} = 729

(\sqrt{3})^n = 729 = 3^6

3^{n/2} = 3^6

\frac{n}{2} = 6 \Rightarrow n = 12

729 is the $12^{\text{th}}$ term.

---

(c)

\dfrac{1}{3}, \dfrac{1}{9}, \dfrac{1}{27}, \ldots

First term

a = \dfrac{1}{3}

, common ratio

r = \dfrac{1/9}{1/3} = \dfrac{1}{3}

.

Let

a_n = \dfrac{1}{19683}

\frac{1}{3} \cdot \left(\frac{1}{3}\right)^{n-1} = \frac{1}{19683}

\left(\frac{1}{3}\right)^n = \frac{1}{3^9} \quad (\text{since } 19683 = 3^9)

3^{-n} = 3^{-9} \Rightarrow n = 9

$\dfrac{1}{19683}$ is the $9^{\text{th}}$ term.

6For what values of

x

, the numbers

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

are in G.P.?Show solution

Condition for G.P.: The middle term squared equals the product of the other two terms.

x^2 = \left(-\frac{2}{7}\right)\left(-\frac{7}{2}\right)

x^2 = \frac{2}{7} \times \frac{7}{2} = 1

x = \pm 1

Therefore, $x = 1$ or $x = -1$ .

7Find the sum to 20 terms of the G.P.:

0.15, 0.015, 0.0015, \ldots

Show solution

Given:

a = 0.15

r = \dfrac{0.015}{0.15} = 0.1 = \dfrac{1}{10}

n = 20

.

Formula:

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

(since

|r| &lt; 1

)

S_{20} = \frac{0.15\left(1 - \left(\dfrac{1}{10}\right)^{20}\right)}{1 - \dfrac{1}{10}} = \frac{0.15\left(1 - 10^{-20}\right)}{\dfrac{9}{10}}

= \frac{0.15 \times 10}{9}\left(1 - 10^{-20}\right) = \frac{1.5}{9}\left(1 - 10^{-20}\right)

= \frac{1}{6}\left(1 - 10^{-20}\right) = \frac{1}{6}\left(1 - \frac{1}{10^{20}}\right)

Therefore, $S_{20} = \dfrac{1}{6}\left(1 - \dfrac{1}{10^{20}}\right)$ .

8Find the sum to

n

terms of the G.P.:

\sqrt{7}, \sqrt{21}, 3\sqrt{7}, \ldots

Show solution

Given:

a = \sqrt{7}

Common ratio:

r = \dfrac{\sqrt{21}}{\sqrt{7}} = \sqrt{\dfrac{21}{7}} = \sqrt{3}

Formula:

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

(since

r &gt; 1

)

S_n = \frac{\sqrt{7}\left((\sqrt{3})^n - 1\right)}{\sqrt{3} - 1}

Rationalising the denominator by multiplying numerator and denominator by

(\sqrt{3}+1)

S_n = \frac{\sqrt{7}\left((\sqrt{3})^n - 1\right)(\sqrt{3}+1)}{(\sqrt{3})^2 - 1^2} = \frac{\sqrt{7}\left((\sqrt{3})^n - 1\right)(\sqrt{3}+1)}{2}

Therefore, $S_n = \dfrac{\sqrt{7}(\sqrt{3}+1)\left[(\sqrt{3})^n - 1\right]}{2}$ .

9Find the sum to

n

terms of the G.P.:

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

(if

a \neq -1

).Show solution

Given G.P.: First term

= 1

, common ratio

= \dfrac{-a}{1} = -a

.

Formula:

S_n = \dfrac{1 \cdot (1 - (-a)^n)}{1 - (-a)} = \dfrac{1 - (-a)^n}{1 + a}

Therefore, $S_n = \dfrac{1 - (-a)^n}{1 + a}$ (valid for

a \neq -1

10Find the sum to

n

terms of the G.P.:

x^3, x^5, x^7, \ldots

(if

x \neq \pm 1

).Show solution

Given G.P.: First term

a = x^3

, common ratio

r = \dfrac{x^5}{x^3} = x^2

.

Formula:

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

S_n = \frac{x^3\left((x^2)^n - 1\right)}{x^2 - 1} = \frac{x^3(x^{2n} - 1)}{x^2 - 1}

Therefore, $S_n = \dfrac{x^3(x^{2n} - 1)}{x^2 - 1}$ (valid for

x \neq \pm 1

11Evaluate

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

.Show solution

Expanding the sum:

\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 (G.P. with

a = 3

r = 3

n = 11

\sum_{k=1}^{11} 3^k = 3 + 3^2 + \ldots + 3^{11} = \frac{3(3^{11} - 1)}{3 - 1} = \frac{3(177147 - 1)}{2} = \frac{3 \times 177146}{2} = \frac{531438}{2} = 265719

Total:

\sum_{k=1}^{11}(2 + 3^k) = 22 + 265719 = 265741

Therefore, the value is $265741$ .

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

\dfrac{39}{10}

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

Let the three terms be

\dfrac{a}{r}, a, ar

.

Product condition:

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

Sum condition:

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

\frac{1 + r + r^2}{r} = \frac{39}{10}

10(1 + r + r^2) = 39r

10r^2 - 29r + 10 = 0

10r^2 - 25r - 4r + 10 = 0

(2r - 5)(5r - 2) = 0

r = \frac{5}{2} \quad \text{or} \quad r = \frac{2}{5}

When $r = \dfrac{5}{2}$ : Terms are

\dfrac{2}{5}, 1, \dfrac{5}{2}

.

When $r = \dfrac{2}{5}$ : Terms are

\dfrac{5}{2}, 1, \dfrac{2}{5}

.

Therefore, the common ratio is $\dfrac{5}{2}$ or $\dfrac{2}{5}$ , and the terms are $\dfrac{2}{5}, 1, \dfrac{5}{2}$ (or in reverse order).

13How many terms of G.P.

3, 3^2, 3^3, \ldots

are needed to give the sum 120?Show solution

Given G.P.:

a = 3

r = 3

.

Sum formula:

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

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

240 = 3(3^n - 1)

80 = 3^n - 1

3^n = 81 = 3^4

n = 4

Therefore, 4 terms are needed.

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 first term

= a

, common ratio

= r

.

Sum of first three terms:

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

Sum of next three terms (4th, 5th, 6th):

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

Dividing (2) by (1):

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

r^3 = 8 \Rightarrow r = 2

Substituting $r = 2$ in (1):

a(1 + 2 + 4) = 16 \Rightarrow 7a = 16 \Rightarrow a = \frac{16}{7}

Sum to $n$ terms:

S_n = \frac{a(r^n - 1)}{r - 1} = \frac{\dfrac{16}{7}(2^n - 1)}{2 - 1} = \frac{16(2^n - 1)}{7}

Therefore, $a = \dfrac{16}{7}$ , $r = 2$ , and $S_n = \dfrac{16(2^n - 1)}{7}$ .

15Given a G.P. with

a = 729

and

7^{\text{th}}

term 64, determine

S_7

.Show solution

Given:

a = 729

a_7 = 64

.

Finding $r$ :

a_7 = ar^6 \Rightarrow 729 \cdot r^6 = 64

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

r = \frac{2}{3}

Sum formula (since

r &lt; 1

S_7 = \frac{a(1 - r^7)}{1 - r} = \frac{729\left(1 - \left(\dfrac{2}{3}\right)^7\right)}{1 - \dfrac{2}{3}}

= \frac{729\left(1 - \dfrac{128}{2187}\right)}{\dfrac{1}{3}} = 729 \times 3 \times \left(\frac{2187 - 128}{2187}\right)

= 2187 \times \frac{2059}{2187} = 2059

Therefore, $S_7 = 2059$ .

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 first term

= a

, common ratio

= r

.

Condition 1:

a_5 = 4 \cdot a_3

ar^4 = 4ar^2 \Rightarrow r^2 = 4 \Rightarrow r = \pm 2

Condition 2:

a_1 + a_2 = -4

a + ar = -4 \Rightarrow a(1 + r) = -4

Case 1: $r = 2$

a(1 + 2) = -4 \Rightarrow a = -\frac{4}{3}

G.P.:

-\dfrac{4}{3},\ -\dfrac{8}{3},\ -\dfrac{16}{3},\ \ldots

Case 2: $r = -2$

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

G.P.:

4,\ -8,\ 16,\ -32,\ \ldots

Therefore, the G.P. is $-\dfrac{4}{3}, -\dfrac{8}{3}, -\dfrac{16}{3}, \ldots$ or $4, -8, 16, -32, \ldots$

17If the

4^{\text{th}}

10^{\text{th}}

and

16^{\text{th}}

terms of a G.P. are

x, y

and

z

, respectively. Prove that

x, y, z

are in G.P.Show solution

Let first term

= a

, common ratio

= r

x = a_4 = ar^3, \quad y = a_{10} = ar^9, \quad z = a_{16} = ar^{15}

Check if $y^2 = xz$ :

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

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

Since $y^2 = xz$ , the numbers $x, y, z$ are in G.P.

\qquad \textbf{(Proved)}

18Find the sum to

n

terms of the sequence

8, 88, 888, 8888, \ldots

Show solution

The general term can be written as:

S_n = 8 + 88 + 888 + \ldots \text{ to } n \text{ terms}

= 8(1 + 11 + 111 + \ldots \text{ to } n \text{ terms})

= \frac{8}{9}(9 + 99 + 999 + \ldots \text{ to } n \text{ terms})

= \frac{8}{9}\left[(10-1) + (10^2-1) + (10^3-1) + \ldots + (10^n - 1)\right]

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

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

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

= \frac{8}{81}\left[10(10^n - 1) - 9n\right]

Therefore, $S_n = \dfrac{8}{81}\left[10^{n+1} - 9n - 10\right]$ .

19Find the sum of the products of the corresponding terms of the sequences

2, 4, 8, 16, 32

and

128, 32, 8, 2, \dfrac{1}{2}

.Show solution

Products of corresponding terms:

2 \times 128,\ 4 \times 32,\ 8 \times 8,\ 16 \times 2,\ 32 \times \frac{1}{2}

= 256,\ 128,\ 64,\ 32,\ 16

This is a G.P. with

a = 256

r = \dfrac{128}{256} = \dfrac{1}{2}

n = 5

S_5 = \frac{256\left(1 - \left(\dfrac{1}{2}\right)^5\right)}{1 - \dfrac{1}{2}} = \frac{256\left(1 - \dfrac{1}{32}\right)}{\dfrac{1}{2}} = 512 \times \frac{31}{32} = 16 \times 31 = 496

Therefore, the required sum is $496$ .

20Show that the products of the corresponding terms of the sequences

a, ar, ar^2, \ldots, ar^{n-1}

and

A, AR, AR^2, \ldots, AR^{n-1}

form a G.P., and find the common ratio.Show solution

Products of corresponding terms:

aA,\ ar \cdot AR,\ ar^2 \cdot AR^2,\ \ldots,\ ar^{n-1} \cdot AR^{n-1}

= aA,\ aArR,\ aAr^2R^2,\ \ldots,\ aA(rR)^{n-1}

Ratio of consecutive terms:

\frac{aA(rR)^k}{aA(rR)^{k-1}} = rR \quad \text{(constant for all } k\text{)}

Since the ratio of consecutive terms is constant ( $= rR$ ), the products form a G.P.

Common ratio $= rR$ .

\qquad \textbf{(Proved)}

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

4^{\text{th}}

by 18.Show solution

Let the four terms be

a, ar, ar^2, ar^3

.

Condition 1:

a_3 - a_1 = 9

ar^2 - a = 9 \Rightarrow a(r^2 - 1) = 9 \quad \ldots (1)

Condition 2:

a_2 - a_4 = 18

ar - ar^3 = 18 \Rightarrow ar(1 - r^2) = 18 \quad \ldots (2)

Dividing (2) by (1):

\frac{ar(1 - r^2)}{a(r^2 - 1)} = \frac{18}{9}

\frac{-ar(r^2 - 1)}{a(r^2 - 1)} = 2

-r = 2 \Rightarrow r = -2

Substituting $r = -2$ in (1):

a(4 - 1) = 9 \Rightarrow 3a = 9 \Rightarrow a = 3

The four terms are:

3,\ 3(-2),\ 3(-2)^2,\ 3(-2)^3 = 3,\ -6,\ 12,\ -24

Therefore, the four numbers are $3, -6, 12, -24$ .

22If the

p^{\text{th}}

q^{\text{th}}

and

r^{\text{th}}

terms of a G.P. are

a, b

and

c

, respectively. Prove that

a^{q-r} b^{r-p} c^{p-q} = 1

.Show solution

Let first term

= A

, common ratio

= R

a = AR^{p-1}, \quad b = AR^{q-1}, \quad c = AR^{r-1}

Now compute $a^{q-r} b^{r-p} c^{p-q}$ :

= (AR^{p-1})^{q-r} \cdot (AR^{q-1})^{r-p} \cdot (AR^{r-1})^{p-q}

= A^{(q-r)+(r-p)+(p-q)} \cdot R^{(p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q)}

Exponent of $A$ :

(q - r) + (r - p) + (p - q) = 0

Exponent of $R$ :

(p-1)(q-r) + (q-1)(r-p) + (r-1)(p-q)

= p(q-r) + q(r-p) + r(p-q) - [(q-r)+(r-p)+(p-q)]

= pq - pr + qr - qp + rp - rq - 0 = 0

Therefore:

a^{q-r} b^{r-p} c^{p-q} = A^0 \cdot R^0 = 1 \qquad \textbf{(Proved)}

23If the first and the

n^{\text{th}}

term of a G.P. are

a

and

b

, respectively, and if

P

is the product of

n

terms, prove that

P^2 = (ab)^n

.Show solution

Let first term

= a

, common ratio

= r

.

The $n^{\text{th}}$ term:

b = ar^{n-1}

, so

r^{n-1} = \dfrac{b}{a}

.

Product of $n$ terms:

P = a \cdot ar \cdot ar^2 \cdots ar^{n-1} = a^n \cdot r^{0+1+2+\cdots+(n-1)} = a^n \cdot r^{n(n-1)/2}

Therefore:

P^2 = a^{2n} \cdot r^{n(n-1)}

Now compute $(ab)^n$ :

ab = a \cdot ar^{n-1} = a^2 r^{n-1}

(ab)^n = a^{2n} r^{n(n-1)}

Hence $P^2 = (ab)^n$ .

\qquad \textbf{(Proved)}

24Show that the ratio of the sum of first

n

terms of a G.P. to the sum of terms from

(n+1)^{\text{th}}

(2n)^{\text{th}}

term is

\dfrac{1}{r^n}

.Show solution

Let first term

= a

, common ratio

= r

(

r \neq 1

).

Sum of first $n$ terms:

S_1 = \frac{a(r^n - 1)}{r - 1}

Sum of first $2n$ terms:

S_2 = \frac{a(r^{2n} - 1)}{r - 1}

Sum of terms from $(n+1)^{\text{th}}$ to $(2n)^{\text{th}}$ :

S_2 - S_1 = \frac{a(r^{2n} - 1) - a(r^n - 1)}{r - 1} = \frac{a(r^{2n} - r^n)}{r - 1} = \frac{ar^n(r^n - 1)}{r - 1}

Required ratio:

\frac{S_1}{S_2 - S_1} = \frac{\dfrac{a(r^n - 1)}{r-1}}{\dfrac{ar^n(r^n - 1)}{r-1}} = \frac{a(r^n-1)}{ar^n(r^n-1)} = \frac{1}{r^n}

Hence proved.

\qquad \textbf{(Proved)}

25If

a, b, c

and

d

are in G.P., show that

(a^2 + b^2 + c^2)(b^2 + c^2 + d^2) = (ab + bc + cd)^2

.Show solution

Since $a, b, c, d$ are in G.P., let common ratio

= r

. Then

b = ar

c = ar^2

d = ar^3

.

LHS:

a^2 + b^2 + c^2 = a^2(1 + r^2 + r^4)

b^2 + c^2 + d^2 = a^2r^2(1 + r^2 + r^4)

\text{LHS} = a^2(1+r^2+r^4) \cdot a^2r^2(1+r^2+r^4) = a^4r^2(1+r^2+r^4)^2

RHS:

ab + bc + cd = a \cdot ar + ar \cdot ar^2 + ar^2 \cdot ar^3 = a^2r + a^2r^3 + a^2r^5 = a^2r(1 + r^2 + r^4)

(ab+bc+cd)^2 = a^4r^2(1+r^2+r^4)^2

LHS = RHS.

\qquad \textbf{(Proved)}

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

Let the two numbers to be inserted be

G_1

and

G_2

, so the sequence is

3, G_1, G_2, 81

.

This is a G.P. with first term

a = 3

, fourth term

= 81

, and

n = 4

a_4 = ar^3 \Rightarrow 3r^3 = 81 \Rightarrow r^3 = 27 \Rightarrow r = 3

G_1 = ar = 3 \times 3 = 9

G_2 = ar^2 = 3 \times 9 = 27

Therefore, the two numbers to be inserted are $9$ and $27$ .

27Find the value of

n

so that

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

may be the geometric mean between

a

and

b

.Show solution

The geometric mean (G.M.) between $a$ and $b$ is $\sqrt{ab}$ .

Setting the expression equal to G.M.:

\frac{a^{n+1} + b^{n+1}}{a^n + b^n} = \sqrt{ab}

a^{n+1} + b^{n+1} = \sqrt{ab}(a^n + b^n)

a^{n+1} + b^{n+1} = a^{1/2}b^{1/2}(a^n + b^n)

a^{n+1} + b^{n+1} = a^{n + 1/2}b^{1/2} + a^{1/2}b^{n+1/2}

a^{n+1} - a^{n+1/2}b^{1/2} = a^{1/2}b^{n+1/2} - b^{n+1}

a^{n+1/2}(a^{1/2} - b^{1/2}) = b^{n+1/2}(a^{1/2} - b^{1/2})

Since

a \neq b

, we can divide both sides by

(a^{1/2} - b^{1/2})

a^{n+1/2} = b^{n+1/2}

\left(\frac{a}{b}\right)^{n+1/2} = 1

Since

a \neq b

, this requires

n + \dfrac{1}{2} = 0

, i.e.,

n = -\dfrac{1}{2}

.

Therefore, $n = -\dfrac{1}{2}$ .

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

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

.Show solution

Let the two numbers be

a

and

b

. Their G.M.

= \sqrt{ab}

.

Given:

a + b = 6\sqrt{ab}

Using componendo-dividendo:

\frac{a + b}{2\sqrt{ab}} = \frac{6\sqrt{ab}}{2\sqrt{ab}} = 3

Let

\dfrac{a}{b} = k

. Then:

\frac{a + b}{2\sqrt{ab}} = \frac{\sqrt{a/b} + \sqrt{b/a}}{2} = \frac{\sqrt{k} + 1/\sqrt{k}}{2} = 3

\sqrt{k} + \frac{1}{\sqrt{k}} = 6

Let

t = \sqrt{k}

t + \frac{1}{t} = 6 \Rightarrow t^2 - 6t + 1 = 0

t = \frac{6 \pm \sqrt{36 - 4}}{2} = \frac{6 \pm \sqrt{32}}{2} = 3 \pm 2\sqrt{2}

\sqrt{k} = 3 + 2\sqrt{2} \Rightarrow k = (3 + 2\sqrt{2})^2

\frac{a}{b} = (3 + 2\sqrt{2})^2 = \frac{(3+2\sqrt{2})^2}{1}

Alternatively, using componendo-dividendo on

a + b = 6\sqrt{ab}

\frac{(a+b)^2}{4ab} = 9 \Rightarrow (a+b)^2 = 36ab

(a - b)^2 = (a+b)^2 - 4ab = 36ab - 4ab = 32ab

\frac{a+b}{a-b} = \frac{6\sqrt{ab}}{\sqrt{32ab}} = \frac{6}{4\sqrt{2}} = \frac{3}{2\sqrt{2}}

Applying componendo-dividendo:

\frac{a}{b} = \frac{3 + 2\sqrt{2}}{3 - 2\sqrt{2}}

Hence, $a : b = (3 + 2\sqrt{2}) : (3 - 2\sqrt{2})$ .

\qquad \textbf{(Proved)}

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)}

.Show solution

Let the two positive numbers be

a

and

b

A = \frac{a+b}{2} \Rightarrow a + b = 2A \quad \ldots (1)

G = \sqrt{ab} \Rightarrow ab = G^2 \quad \ldots (2)

The numbers $a$ and $b$ are roots of:

t^2 - (a+b)t + ab = 0

t^2 - 2At + G^2 = 0

t = \frac{2A \pm \sqrt{4A^2 - 4G^2}}{2} = A \pm \sqrt{A^2 - G^2} = A \pm \sqrt{(A+G)(A-G)}

Therefore, the two numbers are $A + \sqrt{(A+G)(A-G)}$ and $A - \sqrt{(A+G)(A-G)}$ .

\qquad \textbf{(Proved)}

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

2^{\text{nd}}

hour,

4^{\text{th}}

hour and

n^{\text{th}}

hour?Show solution

Given: Initial bacteria

= 30

, doubles every hour, so

r = 2

.

The number of bacteria at the end of

k^{\text{th}}

hour

= 30 \times 2^k

.

At end of $2^{\text{nd}}$ hour:

= 30 \times 2^2 = 30 \times 4 = 120

At end of $4^{\text{th}}$ hour:

= 30 \times 2^4 = 30 \times 16 = 480

At end of $n^{\text{th}}$ hour:

= 30 \times 2^n

Therefore, the number of bacteria at the end of $2^{\text{nd}}$ , $4^{\text{th}}$ and $n^{\text{th}}$ hour are $120$ , $480$ and $30 \times 2^n$ respectively.

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

Given: Principal

P = 500

, rate

r = 10\% = 0.10

, time

= 10

years.

Formula for compound interest:

A = P\left(1 + \frac{r}{100}\right)^n = 500\left(1 + \frac{10}{100}\right)^{10} = 500(1.1)^{10}

Therefore, the amount after 10 years is $500(1.1)^{10}$ rupees.

(Using

(1.1)^{10} \approx 2.5937

, Amount

\approx

1296.87

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

Let the roots be

\alpha

and

\beta

.

A.M.:

\dfrac{\alpha + \beta}{2} = 8 \Rightarrow \alpha + \beta = 16

G.M.:

\sqrt{\alpha\beta} = 5 \Rightarrow \alpha\beta = 25

Quadratic equation with roots

\alpha

and

\beta

x^2 - (\alpha + \beta)x + \alpha\beta = 0

x^2 - 16x + 25 = 0

Therefore, the required quadratic equation is $x^2 - 16x + 25 = 0$ .

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

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

, find the value of

n

.Show solution

Given:

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

f(1) = 3

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

.

Finding $f(x)$ :

Put

y = 1

f(x+1) = f(x) \cdot f(1) = 3f(x)

This means

f(x)

is a G.P. with first term

f(1) = 3

and common ratio

3

f(x) = 3^x

Sum:

\sum_{x=1}^{n} f(x) = 3 + 3^2 + 3^3 + \ldots + 3^n = \frac{3(3^n - 1)}{3 - 1} = \frac{3(3^n - 1)}{2} = 120

3(3^n - 1) = 240

3^n - 1 = 80

3^n = 81 = 3^4

n = 4

Therefore, $n = 4$ .

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.Show solution

Given:

a = 5

r = 2

S_n = 315

S_n = \frac{a(r^n - 1)}{r - 1} = \frac{5(2^n - 1)}{1} = 5(2^n - 1) = 315

2^n - 1 = 63 \Rightarrow 2^n = 64 = 2^6 \Rightarrow n = 6

Last term:

a_6 = ar^{n-1} = 5 \times 2^5 = 5 \times 32 = 160

Therefore, the number of terms is $6$ and the last term is $160$ .

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.Show solution

Given:

a = 1

a_3 + a_5 = 90

a_3 = ar^2 = r^2, \quad a_5 = ar^4 = r^4

r^2 + r^4 = 90

r^4 + r^2 - 90 = 0

Let

u = r^2

u^2 + u - 90 = 0

(u + 10)(u - 9) = 0

u = 9 \quad (\text{since } u = r^2 \geq 0, \text{ reject } u = -10)

r^2 = 9 \Rightarrow r = \pm 3

Therefore, the common ratio is $r = 3$ or $r = -3$ .

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.Show solution

Let the three numbers in G.P. be

a, ar, ar^2

.

Condition 1:

a + ar + ar^2 = 56

a(1 + r + r^2) = 56 \quad \ldots (1)

Condition 2:

a-1,\ ar-7,\ ar^2-21

are in A.P.

2(ar - 7) = (a - 1) + (ar^2 - 21)

2ar - 14 = a + ar^2 - 22

2ar - a - ar^2 = -8

a(2r - 1 - r^2) = -8

-a(r^2 - 2r + 1) = -8

a(r-1)^2 = 8 \quad \ldots (2)

Dividing (1) by (2):

\frac{1 + r + r^2}{(r-1)^2} = 7

1 + r + r^2 = 7(r^2 - 2r + 1)

1 + r + r^2 = 7r^2 - 14r + 7

6r^2 - 15r + 6 = 0

2r^2 - 5r + 2 = 0

(2r - 1)(r - 2) = 0

r = \frac{1}{2} \quad \text{or} \quad r = 2

When $r = 2$ : From (2):

a(1)^2 = 8 \Rightarrow a = 8

. Terms:

8, 16, 32

.

When $r = \dfrac{1}{2}$ : From (2):

a\left(-\dfrac{1}{2}\right)^2 = 8 \Rightarrow a = 32

. Terms:

32, 16, 8

.

Therefore, the three numbers are $8, 16, 32$ .

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.Show solution

Let the G.P. have

2n

terms with first term

a

and common ratio

r

.

Sum of all $2n$ terms:

S = \frac{a(r^{2n} - 1)}{r - 1}

Terms at odd places:

a_1, a_3, a_5, \ldots, a_{2n-1}

— these form a G.P. with first term

a

and common ratio

r^2

, having

n

terms.

Sum of terms at odd places:

S_{\text{odd}} = \frac{a(r^{2n} - 1)}{r^2 - 1} = \frac{a(r^{2n}-1)}{(r-1)(r+1)}

Given:

S = 5 S_{\text{odd}}

\frac{a(r^{2n}-1)}{r-1} = 5 \cdot \frac{a(r^{2n}-1)}{(r-1)(r+1)}

1 = \frac{5}{r+1}

r + 1 = 5 \Rightarrow r = 4

Therefore, the common ratio is $4$ .

6If

\dfrac{a+bx}{a-bx} = \dfrac{b+cx}{b-cx} = \dfrac{c+dx}{c-dx}

(x \neq 0)

, then show that

a, b, c

and

d

are in G.P.Show solution

Let

\dfrac{a+bx}{a-bx} = \dfrac{b+cx}{b-cx} = k

.

Using componendo-dividendo on

\dfrac{a+bx}{a-bx} = \dfrac{b+cx}{b-cx}

\frac{(a+bx)+(a-bx)}{(a+bx)-(a-bx)} = \frac{(b+cx)+(b-cx)}{(b+cx)-(b-cx)}

\frac{2a}{2bx} = \frac{2b}{2cx}

\frac{a}{b} = \frac{b}{c} \Rightarrow b^2 = ac \quad \ldots (1)

Similarly, applying componendo-dividendo on

\dfrac{b+cx}{b-cx} = \dfrac{c+dx}{c-dx}

\frac{2b}{2cx} = \frac{2c}{2dx}

\frac{b}{c} = \frac{c}{d} \Rightarrow c^2 = bd \quad \ldots (2)

From (1) and (2):

\dfrac{b}{a} = \dfrac{c}{b} = \dfrac{d}{c}

, which means

a, b, c, d

are in G.P.

\qquad \textbf{(Proved)}

7Let

S

be the sum,

P

the product and

R

the sum of reciprocals of

n

terms in a G.P. Prove that

P^2 R^n = S^n

.Show solution

Let the G.P. have first term

a

and common ratio

r

.

Sum:

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

Product:

P = a^n \cdot r^{0+1+\cdots+(n-1)} = a^n r^{n(n-1)/2}

Sum of reciprocals (the reciprocals form a G.P. with first term

\dfrac{1}{a}

and ratio

\dfrac{1}{r}

R = \frac{\dfrac{1}{a}\left(1 - \dfrac{1}{r^n}\right)}{1 - \dfrac{1}{r}} = \frac{\dfrac{1}{a} \cdot \dfrac{r^n - 1}{r^n}}{\dfrac{r-1}{r}} = \frac{r^n - 1}{ar^{n-1}(r-1)}

Now compute $P^2 R^n$ :

P^2 = a^{2n} r^{n(n-1)}

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

P^2 R^n = a^{2n} r^{n(n-1)} \cdot \frac{(r^n-1)^n}{a^n r^{n(n-1)}(r-1)^n} = \frac{a^n(r^n-1)^n}{(r-1)^n} = \left[\frac{a(r^n-1)}{r-1}\right]^n = S^n

Hence $P^2 R^n = S^n$ .

\qquad \textbf{(Proved)}

8If

a, b, c, d

are in G.P., prove that

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

are in G.P.Show solution

Note: The question uses

a^n, b^n, c^n, d^n

where

n

is a general exponent (not the number of terms).

Since $a, b, c, d$ are in G.P., let common ratio

= r

. Then

b = ar

c = ar^2

d = ar^3

.

We need to show:

(b^n + c^n)^2 = (a^n + b^n)(c^n + d^n)

LHS:

(b^n + c^n)^2 = (a^n r^n + a^n r^{2n})^2 = a^{2n} r^{2n}(1 + r^n)^2

RHS:

(a^n + b^n)(c^n + d^n) = (a^n + a^n r^n)(a^n r^{2n} + a^n r^{3n})

= a^n(1 + r^n) \cdot a^n r^{2n}(1 + r^n) = a^{2n} r^{2n}(1 + r^n)^2

LHS = RHS, hence

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

are in G.P.

\qquad \textbf{(Proved)}

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

.Show solution

From the first equation:

a + b = 3

ab = p

.

From the second equation:

c + d = 12

cd = q

.

Since $a, b, c, d$ are in G.P., let

a = A

b = AR

c = AR^2

d = AR^3

a + b = A(1 + R) = 3 \quad \ldots (1)

c + d = AR^2(1 + R) = 12 \quad \ldots (2)

Dividing (2) by (1):

R^2 = 4 \Rightarrow R = 2

(taking positive value).

From (1):

A(1 + 2) = 3 \Rightarrow A = 1

.

So

a = 1, b = 2, c = 4, d = 8

p = ab = 1 \times 2 = 2, \quad q = cd = 4 \times 8 = 32

\frac{q + p}{q - p} = \frac{32 + 2}{32 - 2} = \frac{34}{30} = \frac{17}{15}

Hence $(q+p):(q-p) = 17:15$ .

\qquad \textbf{(Proved)}

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

a

and

b

m:n

. Show that

a:b = \left(m + \sqrt{m^2 - n^2}\right) : \left(m - \sqrt{m^2 - n^2}\right)

.Show solution

Given:

\dfrac{\text{A.M.}}{\text{G.M.}} = \dfrac{m}{n}

, i.e.,

\dfrac{(a+b)/2}{\sqrt{ab}} = \dfrac{m}{n}

\frac{a + b}{2\sqrt{ab}} = \frac{m}{n}

Applying componendo-dividendo:

\frac{(a+b) + 2\sqrt{ab}}{(a+b) - 2\sqrt{ab}} = \frac{m+n}{m-n}

\frac{(\sqrt{a} + \sqrt{b})^2}{(\sqrt{a} - \sqrt{b})^2} = \frac{m+n}{m-n}

\frac{\sqrt{a} + \sqrt{b}}{\sqrt{a} - \sqrt{b}} = \sqrt{\frac{m+n}{m-n}}

Applying componendo-dividendo again:

\frac{2\sqrt{a}}{2\sqrt{b}} = \frac{\sqrt{m+n} + \sqrt{m-n}}{\sqrt{m+n} - \sqrt{m-n}}

\frac{a}{b} = \left(\frac{\sqrt{m+n} + \sqrt{m-n}}{\sqrt{m+n} - \sqrt{m-n}}\right)^2 = \frac{(m+n) + (m-n) + 2\sqrt{m^2-n^2}}{(m+n)+(m-n)-2\sqrt{m^2-n^2}}

= \frac{2m + 2\sqrt{m^2-n^2}}{2m - 2\sqrt{m^2-n^2}} = \frac{m + \sqrt{m^2-n^2}}{m - \sqrt{m^2-n^2}}

Hence $a:b = \left(m + \sqrt{m^2-n^2}\right) : \left(m - \sqrt{m^2-n^2}\right)$ .

\qquad \textbf{(Proved)}

11Find the sum of the following series up to

n

terms: (i)

5 + 55 + 555 + \ldots

(ii)

.6 + .66 + .666 + \ldots

Show solution

(i) $5 + 55 + 555 + \ldots$

S_n = 5 + 55 + 555 + \ldots \text{ to } n \text{ terms}

= 5(1 + 11 + 111 + \ldots)

= \frac{5}{9}(9 + 99 + 999 + \ldots)

= \frac{5}{9}\left[(10-1) + (10^2-1) + \ldots + (10^n - 1)\right]

= \frac{5}{9}\left[\frac{10(10^n - 1)}{9} - n\right]

= \frac{5}{9}\left[\frac{10(10^n-1)}{9} - n\right] = \frac{5}{81}\left[10^{n+1} - 9n - 10\right]

Therefore, $S_n = \dfrac{5}{81}(10^{n+1} - 9n - 10)$ .

---

(ii) $.6 + .66 + .666 + \ldots$

S_n = 0.6 + 0.66 + 0.666 + \ldots \text{ to } n \text{ terms}

= 6(0.1 + 0.11 + 0.111 + \ldots)

= \frac{6}{9}(0.9 + 0.99 + 0.999 + \ldots)

= \frac{2}{3}\left[\left(1 - \frac{1}{10}\right) + \left(1 - \frac{1}{10^2}\right) + \ldots + \left(1 - \frac{1}{10^n}\right)\right]

= \frac{2}{3}\left[n - \frac{1}{10}\cdot\frac{1 - (1/10)^n}{1 - 1/10}\right]

= \frac{2}{3}\left[n - \frac{1}{9}\left(1 - \frac{1}{10^n}\right)\right]

Therefore, $S_n = \dfrac{2}{3}\left[n - \dfrac{1}{9}\left(1 - \dfrac{1}{10^n}\right)\right]$ .

12Find the

20^{\text{th}}

term of the series

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

terms.Show solution

The $n^{\text{th}}$ term of the series:

First factors:

2, 4, 6, \ldots \Rightarrow 2n

Second factors:

4, 6, 8, \ldots \Rightarrow 2n + 2

a_n = 2n(2n+2) = 4n(n+1) = 4n^2 + 4n

For $n = 20$ :

a_{20} = 4(20)^2 + 4(20) = 4 \times 400 + 80 = 1600 + 80 = 1680

Therefore, the $20^{\text{th}}$ term is $1680$ .

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?Show solution

Balance to be paid

= 12000 - 6000 =

6000

.

Annual instalments: Rs 500 principal + 12% interest on unpaid amount.

Year 1: Unpaid

= 6000

. Instalment

= 500 + 12\% \times 6000 = 500 + 720 =

1220

.

Year 2: Unpaid

= 5500

. Instalment

= 500 + 12\% \times 5500 = 500 + 660 =

1160

.

Year 3: Unpaid

= 5000

. Instalment

= 500 + 600 =

1100

\vdots

Year 12: Unpaid

= 500

. Instalment

= 500 + 12\% \times 500 = 500 + 60 =

560

.

The instalments form an A.P.:

1220, 1160, 1100, \ldots, 560

with

d = -60

n = 12

.

Total of instalments:

S = \frac{12}{2}(1220 + 560) = 6 \times 1780 = \text{Rs } 10680

Total cost of tractor:

= 6000 + 10680 = \text{Rs } 16680

Therefore, the tractor will cost him Rs $16680$ .

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?Show solution

Balance

= 22000 - 4000 =

18000

.

Annual instalments: Rs 1000 principal + 10% interest on unpaid amount.

Year 1: Unpaid

= 18000

. Instalment

= 1000 + 10\% \times 18000 = 1000 + 1800 =

2800

.

Year 2: Unpaid

= 17000

. Instalment

= 1000 + 1700 =

2700

.

Year 3: Unpaid

= 16000

. Instalment

= 1000 + 1600 =

2600

\vdots

Year 18: Unpaid

= 1000

. Instalment

= 1000 + 100 =

1100

.

The instalments form an A.P.:

2800, 2700, 2600, \ldots, 1100

with

d = -100

n = 18

.

Total of instalments:

S = \frac{18}{2}(2800 + 1100) = 9 \times 3900 = \text{Rs } 35100

Total cost:

= 4000 + 35100 = \text{Rs } 39100

Therefore, the scooter will cost him Rs $39100$ .

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.Show solution

Number of letters mailed in each set:

- Set 1: 4 letters
- Set 2:

4 \times 4 = 4^2 = 16

letters
- Set 3:

4^3 = 64

letters
-

\vdots

- Set 8:

4^8

letters

Total letters mailed up to and including the $8^{\text{th}}$ set:

S_8 = 4 + 4^2 + 4^3 + \ldots + 4^8 = \frac{4(4^8 - 1)}{4 - 1} = \frac{4(65536 - 1)}{3} = \frac{4 \times 65535}{3} = \frac{262140}{3} = 87380

Cost per letter

= 50

paise

=

0.50

.

Total postage:

= 87380 \times 0.50 = \text{Rs } 43690

Therefore, the amount spent on postage is Rs $43690$ .

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.Show solution

Given: Principal

P =

10000

, rate

= 5\%

per annum (simple interest).

Simple interest per year

= \dfrac{10000 \times 5}{100} =

500

.

Amount at the end of $n^{\text{th}}$ year

= 10000 + 500n

.

Amount in the $15^{\text{th}}$ year (i.e., at end of 15 years):

= 10000 + 500 \times 15 = 10000 + 7500 = \text{Rs } 17500

Total amount after 20 years:

= 10000 + 500 \times 20 = 10000 + 10000 = \text{Rs } 20000

Therefore, the amount in the $15^{\text{th}}$ year is Rs $17500$ and the total amount after 20 years is Rs $20000$ .

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.Show solution

Given: Initial value

=

15625

, depreciation rate

= 20\%

per year.

Value after each year

=

previous value

\times (1 - 0.20) =

previous value

\times 0.8

.

This forms a G.P. with

a = 15625

and

r = 0.8 = \dfrac{4}{5}

.

Value at end of 5 years:

V_5 = 15625 \times \left(\frac{4}{5}\right)^5 = 15625 \times \frac{1024}{3125} = 5 \times 1024 = 5120

Therefore, the estimated value of the machine at the end of 5 years is Rs $5120$ .

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.Show solution

Let the work be originally planned to be completed in

n

days.

Total work (in worker-days)

= 150n

.

Actual workers each day:

150, 146, 142, \ldots

(A.P. with

a = 150

d = -4

).

The work was completed in

(n + 8)

days.

Total actual work done:

S_{n+8} = \frac{n+8}{2}\left[2 \times 150 + (n+8-1)(-4)\right]

= \frac{n+8}{2}\left[300 - 4(n+7)\right]

= \frac{n+8}{2}\left[300 - 4n - 28\right]

= \frac{n+8}{2}(272 - 4n)

= (n+8)(136 - 2n)

Setting actual work $=$ planned work:

(n+8)(136 - 2n) = 150n

136n - 2n^2 + 1088 - 16n = 150n

-2n^2 + 120n + 1088 = 150n

-2n^2 - 30n + 1088 = 0

n^2 + 15n - 544 = 0

(n + 32)(n - 17) = 0

n = 17 \quad (\text{rejecting } n = -32)

Total days to complete work

= n + 8 = 17 + 8 = 25

.

Therefore, the work was completed in $25$ days.

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