Chapter 11: Geometric Progression

Exercise 11A

Question 1

Show that the progression 2, 6, 18, 54, 162,..... is a G.P.
Write its:

(i) first term

(ii) common ratio

(iii) nth term

(iv) 8th term.

Answer

2, 6, 18, 54, 162,.....

$\Rightarrow \dfrac{18}{6} = \dfrac{6}{2}$ = 3.

Since, ratio between consecutive terms are equal, thus the series is in G.P.

a = 2

r = $\dfrac{6}{2}$ = 3

We know that,

nth term of a G.P. is given by,

⇒ T_n = ar^{n - 1}

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

⇒ T₈ = (2)(3)^{8 - 1}

= 2(3)⁷

= 2(2187)

= 4374.

Hence, a = 2, r = 3, T_n = 2.(3)^{n - 1}, T₈ = 4374.

Question 2

Show that the progression 625, 125, 25, 5, 1, $\dfrac{1}{5}$ , ..... is a G.P.
Write its

(i) first term

(ii) common ratio

(iii) nth term

(iv) 10th term

Answer

Given,

625, 125, 25, 5, 1, $\dfrac{1}{5}$ , ........

$\Rightarrow \dfrac{125}{625} = \dfrac{25}{125} = \dfrac{1}{5}.$

Since, ratio between consecutive terms are equal, thus the series is in G.P.

a = 625

r = $\dfrac{125}{625} = \dfrac{1}{5}$

We know that,

nth term of a G.P. is given by,

T_n = ar^{n - 1}

T_n = $625 \times \Big(\dfrac{1}{5}\Big)^{n - 1}$

= $5^4\Big(\dfrac{1}{5^{n - 1}}\Big)$

= $5^{4 - (n - 1)}$

= 5^{5 - n}

= $\dfrac{1}{5^{n - 5}}$ .

10th term,

T₁₀ = 5^{5 - 10}

= 5^-5

= $\dfrac{1}{5^5}$

= $\dfrac{1}{3125}$ .

Hence, a = 625, r = $\dfrac{1}{5}$ , T_n = $\dfrac{1}{5^{n - 5}}$ , T₁₀ = $\dfrac{1}{3125}$ .

Question 3

Show that the progression -27, 9, -3, 1, $-\dfrac{1}{3}$ , ...... is a G.P.
Write its

(i) first term

(ii) common ratio

(iii) nth term

(iv) 9th term.

Answer

Given,

-27, 9, -3, 1, $-\dfrac{1}{3}$ ,......

$\Rightarrow \dfrac{9}{-27} = \dfrac{-3}{9} = -\dfrac{1}{3}$ .

Since, ratio between consecutive terms are equal, thus the series is in G.P.

a = -27

r = $\dfrac{9}{-27} = -\dfrac{1}{3}$

We know that,

nth term of a G.P. is given by,

T_n = ar^{n - 1}

$\Rightarrow T_n = -27 \times \Big(-\dfrac{1}{3}\Big)^{n - 1} \\[1em] = (-3)^3 \times \Big(-\dfrac{1}{3}\Big)^{n - 1} \\[1em] = (-3)^3 \times \dfrac{1}{(-3)^{n - 1}} \\[1em] = (-3)^{3 -(n - 1)} \\[1em] = (-3)^{3 -n + 1} \\[1em] = (-3)^{4 - n} \\[1em] = \dfrac{1}{(-3)^{n - 4}}.$

9th term

$\Rightarrow T_9 = \dfrac{1}{-3^{9 - 4}} \\[1em] = \dfrac{1}{-3^{5}} \\[1em] = -\dfrac{1}{243} \\[1em]$

Hence, a = -27, r = $-\dfrac{1}{3}$ , T_n = = $\dfrac{1}{(-3)^{n - 4}}$ , T₉ = $-\dfrac{1}{243}$ .

Question 4

Show that the progression 2, $2\sqrt{2}, 4, 4\sqrt{2}$ ,..... is a G.P.
Write its

(i) first term

(ii) common ratio

(iii) nth term

(iv) 11th term.

Answer

Given,

2, $2\sqrt{2}, 4, 4\sqrt{2}$ ,.....

$\Rightarrow \dfrac{2\sqrt{2}}{2} = \dfrac{4\sqrt{2}}{4} = \sqrt{2}.$

Since, ratio between consecutive terms are equal, thus the series is in G.P.

a = 2

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

We know that,

nth term of a G.P. is given by,

$\Rightarrow T_n = ar^{n - 1} \\[1em] \Rightarrow T_n = 2.( \sqrt {2})^{n - 1} \\[1em] = (\sqrt{2})^2.(\sqrt {2})^{n - 1} \\[1em] = (\sqrt{2})^{2 + n - 1} \\[1em] = (\sqrt2)^{n + 1}. \\[1em] \Rightarrow T_{11} = (\sqrt2)^{11 + 1} \\[1em] = (\sqrt2)^{12} \\[1em] = (2)^{\dfrac{1}{2} \times 12} \\[1em] = 2^{6} \\[1em] = 64.$

Hence, a = 2, r = $\sqrt{2}$ , T_n = $(\sqrt{2})^{n + 1}$ , T₁₁ = 64.

Question 5

Show that the progression $-\dfrac{3}{4}, \dfrac{1}{2}, -\dfrac{1}{3}, \dfrac{2}{9}$ ,..... is a G.P.
Write its

(i) first term

(ii) common ratio

(iii) nth term

(iv) 6th term

Answer

Given,

$-\dfrac{3}{4}, \dfrac{1}{2}, -\dfrac{1}{3}, \dfrac{2}{9}$ ,.....

$\Rightarrow \dfrac{\dfrac{1}{2}}{-\dfrac{3}{4}} = \dfrac{-\dfrac{1}{3}}{\dfrac{1}{2}} = -\dfrac{2}{3}.$

Since, ratio between consecutive terms are equal, thus the series is in G.P.

a = $-\dfrac{3}{4}$

r = $\dfrac{\dfrac{1}{2}}{-\dfrac{3}{4}} = \dfrac{-4}{6} = \dfrac{-2}{3}$ .

We know that,

nth term of a G.P. is given by,

T_n = ar^{n - 1}

$\Rightarrow T_n = -\dfrac{3}{4} \cdot \Big(-\dfrac{2}{3}\Big)^{n - 1} \\[1em] = \dfrac{-3}{2^2} \cdot \Big(\dfrac{2^{n - 1}}{(-3)^{n - 1}}\Big) \\[1em] = \Big(\dfrac{2^{n - 1 - 2}}{(-3)^{n - 1 - 1}}\Big) \\[1em] = \Big(\dfrac{2^{n - 3}}{(-3)^{n - 2}}\Big)$

6th term,

$T_6 = \Big(\dfrac{2^{6 - 3}}{(-3)^{6 - 2}}\Big) \\[1em] = \Big(\dfrac{2^3}{(-3)^4}\Big) \\[1em] = \dfrac{8}{81}.$

Hence, a = $-\dfrac{3}{4}$ , r = $-\dfrac{2}{3}$ , T_n = $\Big(\dfrac{2^{n - 3}}{(-3)^{n - 2}}\Big)$ , T₆ = $\dfrac{8}{81}$ .

Question 6

Show that the progression 0.4, 0.8, 1.6,..... is a G.P.
Write its

(i) first term

(ii) common ratio

(iii) nth term

(iv) 7th term.

Answer

Given,

0.4, 0.8, 1.6,.....

$\Rightarrow \dfrac{0.8}{0.4}= {\dfrac{1.6}{0.8}} = 2.$

Since, ratio between consecutive terms are equal, thus the series is in G.P.

a = 0.4

r = $\dfrac{0.8}{0.4}$ = 2

We know that,

nth term of a G.P. is given by,

T_n = ar^{n - 1}

$T_n = 0.4(2)^{n - 1} \\[1em] = \dfrac{4}{10}(2)^{n - 1} \\[1em] = \dfrac{2 \times 2}{10}(2)^{n - 1} \\[1em] = \dfrac{2}{5}(2)^{n - 1} \\[1em] = \dfrac{2^{n + 1 - 1}}{5} \\[1em] = \dfrac{2^{n}}{5}.$

7th term,

T₇ = $\dfrac{2^{7}}{5}$

= $\dfrac{128}{5}$ .

Hence, a = 0.4, r = 2, T_n = $\dfrac{2^{n}}{5}$ , T₇ = $\dfrac{128}{5}$ .

Question 7

Which term of the G.P. 3, 6, 12, 24,..... is 768?

Answer

Given,

The G.P. : 3, 6, 12, 24, .....

a = 3

r = $\dfrac{6}{3} = 2$

Let nth term be 768.

T_n = 768

We know that,

nth term of a G.P. is given by,

T_n = ar^{n - 1}

⇒ 768 = (3)(2)^{n - 1}

⇒ $\dfrac{768}{3}$ = (2)^{n - 1}

⇒ 256 = (2)^{n - 1}

⇒ 2⁸ = (2)^{n - 1}

⇒ 8 = n - 1

⇒ n = 8 + 1

⇒ n = 9.

Hence, 9th term of G.P. is 768.

Question 8

Which term of the G.P. 5, 10, 20, 40,...... is 640?

Answer

Given,

The G.P. : 5, 10, 20, 40,......

a = 5

r = $\dfrac{10}{5}$ = 2

Let nth term be 640.

T_n = 640

We know that,

nth term of a G.P. is given by,

T_n = ar^{n - 1}

⇒ 640 = (5).2^{n - 1}

⇒ $\dfrac{640}{5}$ = 2^{n - 1}

⇒ 128 = 2^{n - 1}

⇒ 2⁷ = 2^{n - 1}

Equating the exponents:

⇒ 7 = n - 1

⇒ n = 7 + 1

⇒ n = 8.

Hence, 8th term of G.P. is 640.

Question 9

Which term of the G.P. $\sqrt{3}, 3, 3\sqrt{3}$ , 9, ...... is 729 ?

Answer

Given,

The G.P. : $\sqrt{3}, 3, 3\sqrt{3}$ , 9,......

a = $\sqrt{3}$

r = $\dfrac{3}{\sqrt3} = \sqrt3$

T_n = 729

We know that,

nth term of a G.P. is given by,

T_n = ar^{n - 1}

$\Rightarrow 729 = \sqrt3 \times (\sqrt3)^{n - 1} \\[1em] \Rightarrow 729 = (\sqrt3)^{1 + n - 1} \\[1em] \Rightarrow 729 = (\sqrt3)^{n} \\[1em] \Rightarrow 3^6 = (3)^{\dfrac{n}{2}} \\[1em] \Rightarrow 6 = \dfrac{n}{2} \\[1em] \Rightarrow n = 6 \times 2 \\[1em] \Rightarrow n = 12.$

Hence, 12th term of G.P. is 729.

Question 10

Find the G.P. whose 5th and 8th terms are 80 and 640 respectively.

Answer

Let a be the first term and r be the common ratio.

We know that,

nth term of a G.P. is given by,

T_n = ar^{n - 1}

Given,

5th term = 80

⇒ ar^{5 - 1} = 80

⇒ ar⁴ = 80 ....(1)

Given,

8th term = 640

⇒ ar^{8 - 1} = 640

⇒ ar⁷ = 640 .......(2)

Dividing Equation (2) by Equation (1) :

$\Rightarrow \dfrac{ar^7}{ar^4} = \dfrac{640}{80} \\[1em] \Rightarrow r^{7 - 4} = 8 \\[1em] \Rightarrow r^{3} = 8 \\[1em] \Rightarrow r = \sqrt[3]8 \\[1em] \Rightarrow r = 2.$

Substituting r = 2 in Equation (1), we get :

⇒ a(2)⁴ = 80

⇒ 16a = 80

⇒ a = $\dfrac{80}{16}$

⇒ a = 5.

G.P. is,

5, 5(2), 5(2)², 5(2)³.....

5, 10, 20, 40, .....

Hence, the G.P. is 5, 10, 20, 40, ......

Question 11

The 4th term of a G.P. is 16 and the 7th term is 128. Find the first term and common ratio of the series.

Answer

Let first term be a and common ratio be r.

We know that,

nth term of a G.P. is given by,

T_n = ar^{n - 1}

Given,

4th term of a G.P. is 16.

⇒ ar^{4 - 1} = 16

⇒ ar³ = 16 ......(1)

Given,

7th term is 128.

⇒ ar^{7 - 1} = 128

⇒ ar⁶ = 128 ......(2)

Dividing Equation (2) by Equation (1) :

$\Rightarrow \dfrac{ar^6}{ar^3} = \dfrac{128}{16} \\[1em] \Rightarrow r^{6 - 3} = 8 \\[1em] \Rightarrow r^3 = 8 \\[1em] \Rightarrow r = \sqrt[3]8 \\[1em] \Rightarrow r = 2.$

Substituting r = 2 in Equation 1, we get :

⇒ a(2)³ = 16

⇒ a(8) = 16

⇒ a = $\dfrac{16}{8}$

⇒ a = 2.

Hence, a = 2 and r = 2.

Question 12

Find the G.P. whose 4th and 7th terms are $\dfrac{1}{18}$ and $-\dfrac{1}{486}$ respectively.

Answer

Let a be the first term and r be the common ratio.

We know that,

nth term of a G.P. is given by,

T_n = ar^{n - 1}

Given,

⇒ 4th term of G.P is $\dfrac{1}{18}$

⇒ ar^{4 - 1} = $\dfrac{1}{18}$

⇒ ar³ = $\dfrac{1}{18}$ ....(1)

Given,

⇒ 7th term of G.P. is $-\dfrac{1}{486}$

⇒ ar^{7 - 1} = $-\dfrac{1}{486}$

⇒ ar⁶ = $-\dfrac{1}{486}$ ....(2)

Divide Equation 2 by Equation 1:

$\Rightarrow \dfrac{ar^6}{ar^3} = \dfrac{-\dfrac{1}{486}}{\dfrac{1}{18}} \\[1em] \Rightarrow r^{6 - 3} = -\dfrac{1}{486} \times 18 \\[1em] \Rightarrow r^{3} = -\dfrac{1}{486} \times 18 \\[1em] \Rightarrow r^{3} = -\dfrac{1}{27} \\[1em] \Rightarrow r = \sqrt[3]{-\dfrac{1}{27}} \\[1em] \Rightarrow r = -\dfrac{1}{3}.$

Substituting $r = -\dfrac{1}{3}$ into Equation 1, we get:

$\Rightarrow a\Big(-\dfrac{1}{3}\Big)^3 = \dfrac{1}{18} \\[1em] \Rightarrow a\Big(-\dfrac{1}{27}\Big) = \dfrac{1}{18} \\[1em] \Rightarrow a = -27 \times \dfrac{1}{18} \\[1em] \Rightarrow a = -\dfrac{3}{2}.$

G.P. is,

$\Rightarrow -\dfrac{3}{2}, -\dfrac{3}{2} \times \Big(-\dfrac{1}{3}\Big), -\dfrac{3}{2} \times \Big(-\dfrac{1}{3}\Big)^2, ...... \\[1em] \Rightarrow -\dfrac{3}{2}, \dfrac{1}{2}, -\dfrac{1}{6}, ........$

Hence, the G.P. is $-\dfrac{3}{2}, \dfrac{1}{2}, -\dfrac{1}{6}, ........$

Question 13

For what values of x, the numbers (x + 9), (x - 6) and 4 are in G.P?

Answer

We know that,

In G.P. we have constant common ratio.

$\dfrac{x - 6}{x + 9} = \dfrac{4}{x - 6}$

Solving,

⇒ (x - 6)² = 4(x + 9)

⇒ x² - 12x + 36 = 4x + 36

⇒ x² - 12x + 36 - 4x - 36 = 0

⇒ x² - 16x = 0

⇒ x(x - 16) = 0

⇒ x = 0 or (x - 16) = 0

⇒ x = 0 or x = 16.

Hence, x = 0 or x = 16.

Question 14

Find the 6th term from the end of the G.P. : 16, 8, 4, 2,......, $\dfrac{1}{512}$ .

Answer

Given,

a = 16

r = $\dfrac{8}{16} = \dfrac{1}{2}$

l = $\dfrac{1}{512}$

We know that,

nth term from end of a G.P. is given by,

T_n = $\dfrac{l}{r^{n - 1}}$

$\Rightarrow T_{\text{6th term from end}}= \dfrac{\dfrac{1}{512}}{\Big(\dfrac{1}{2}\Big)^{6 - 1}} \\[1em] \Rightarrow \dfrac{\dfrac{1}{512}}{\Big(\dfrac{1}{2}\Big)^{5}} \\[1em] \Rightarrow \dfrac{\dfrac{1}{512}}{\Big(\dfrac{1}{32}\Big)} \\[1em] \Rightarrow \dfrac{1}{512} \times 32 \\[1em] \Rightarrow \dfrac{1}{16}.$

Hence, 6th term from end = $\dfrac{1}{16}$ .

Question 15

Find the 4th term from the end of the G.P. $\dfrac{2}{81},\dfrac{2}{27},\dfrac{2}{9},......,54$ .

Answer

a = $\dfrac{2}{81}$

r = $\dfrac{\dfrac{2}{27}}{\dfrac{2}{81}} = 3$

l = 54

We know that,

nth term from end of a G.P. is given by,

T_n = $\dfrac{l}{r^{n - 1}}$

$\Rightarrow \text{4th from end}= \dfrac{54}{3^{4-1}} \\[1em] = \dfrac{54}{3^{3}} \\[1em] = \dfrac{54}{27} \\[1em] = 2.$

Hence, 4th term from end = 2.

Question 16

The 4th, 6th and last term of a geometric progression are 10, 40 and 640 respectively. If the common ratio is positive, find the first term, common ratio and the number of terms of the series.

Answer

We know that,

nth term of a G.P. is given by,

T_n = ar^{n - 1}

Given,

4th term = 10

⇒ T₄ = ar³

⇒ ar³ = 10 .....(1)

Given,

6th term = 40

⇒ T₆ = ar⁵

⇒ ar⁵ = 40 .....(2)

Given,

Let nth term be last term.

⇒ T_n = ar^{n - 1}

Since, last term = 640

⇒ ar^{n - 1} = 640 .....(3)

Divide Equation 2 by Equation 1:

$\Rightarrow \dfrac{ar^5}{ar^3} = \dfrac{40}{10}$

⇒ r^{5 - 3} = 4

⇒ r² = 4

⇒ r = 2 (Since, common ratio is positive)

Substitute r = 2 into Equation 1:

⇒ a(2)³ = 10

⇒ a(8) = 10

⇒ a = $\dfrac{10}{8}$

⇒ a = $\dfrac{5}{4}$ .

Now, substituting values of a and r in equation (3), we get :

⇒ $\dfrac{5}{4} \times$ 2^{n - 1} = 640

⇒ 2^{n - 1} = $\dfrac{640 × 4}{5}$

⇒ 2^{n - 1} = 128 × 4

⇒ 2^{n - 1} = 512

⇒ 2^{n - 1} = 2⁹

⇒ n - 1 = 9

⇒ n = 9 + 1

⇒ n = 10.

Hence, a = $\dfrac{5}{4}$ , r = 2 and n = 10.

Exercise 11B

Question 1

Find the sum of first 8 terms of the G.P. 1, 3, 9, 27, 81, .....

Answer

Given,

a = 1

r = $\dfrac{3}{1}$ = 3

n = 8

We know that,

The sum of the first n terms of a G.P. is given by :

$S_n = \dfrac{a(r^n - 1)}{r - 1}$ [For r > 1]

Substituting values we get :

$\Rightarrow S_8 = \dfrac{1(3^8 - 1)}{3 - 1} \\[1em] = \dfrac{3^8 - 1}{2} \\[1em] = \dfrac{6561 - 1}{2} \\[1em] = \dfrac{6560}{2} \\[1em] = 3280.$

Hence, S₈ = 3280.

Question 2

Find the sum of first 10 terms of the G.P. 1, $\sqrt{3}$ , 3, $3\sqrt{3}$ , ………

Answer

Given,

a = 1

r = $\dfrac{\sqrt3}{1} = \sqrt3$

n = 10

We know that,

The sum of the first n terms of a G.P. is given by:

$S_n = \dfrac{a(r^n - 1)}{r - 1}$ [For r > 1]

Substituting values we get :

$\Rightarrow S_{10} = \dfrac{1[(\sqrt3)^{10} - 1]}{\sqrt3 - 1} \\[1em] = \dfrac{(3)^{\dfrac{10}{2}} - 1}{\sqrt3 - 1} \\[1em] = \dfrac{(3)^{5} - 1}{\sqrt3 - 1} \\[1em] = \dfrac{243 - 1}{\sqrt3 - 1} \\[1em] = \dfrac{242}{\sqrt3 - 1}$

Rationalizing the Denominator :

$= \dfrac{242}{\sqrt3 - 1} \times \dfrac{\sqrt3 + 1}{\sqrt3 + 1}\\[1em] = \dfrac{242(\sqrt3 + 1)}{(\sqrt3)^2 - 1^2} \\[1em] = \dfrac{242(\sqrt3 + 1)}{3 - 1} \\[1em] = \dfrac{242(\sqrt3 + 1)}{2} \\[1em] = 121(\sqrt3 + 1).$

Hence, S₁₀ = $121(\sqrt3 + 1)$ .

Question 3

Find the sum of first 9 terms of the G.P. 1, $-\dfrac{1}{2}$ , $\dfrac{1}{4}$ , $-\dfrac{1}{8}$ , ………

Answer

Given,

a = 1

r = $\dfrac{\dfrac{-1}{2}}{1} = -\dfrac{1}{2}$

n = 9

We know that,

The sum of the first n terms of a G.P. is given by:

$S_n = \dfrac{a(1 - r^n)}{1 - r}$ [For r < 1]

Substituting values we get :

$\Rightarrow S_9 = \dfrac{1\Big[1 - \Big(\dfrac{-1}{2}\Big)^9\Big]}{1 - \Big(\dfrac{-1}{2}\Big)} \\[1em] = \dfrac{\Big(1 + \dfrac{1}{512}\Big)}{1 + \dfrac{1}{2}} \\[1em] = \dfrac{\dfrac{512 + 1}{512}}{\dfrac{2 + 1}{2}} \\[1em] = \dfrac{\dfrac{513}{512}}{\dfrac{3}{2}} \\[1em] = \dfrac{513}{512} \times {\dfrac{2}{3}} \\[1em] = \dfrac{171}{256}.$

Hence, S₉ = $\dfrac{171}{256}$ .

Question 4

Find the sum of first 6 terms of the G.P. 0.1, 0.01, 0.001, .....

Answer

Given,

a = 0.1

r = $\dfrac{0.01}{0.1}$ = 0.1

n = 6

We know that,

The sum of the first n terms of a G.P. is given by :

$S_n = \dfrac{a(1 - r^n)}{1 - r}$ [For r < 1]

Substituting values we get :

$\Rightarrow S_6 = \dfrac{0.1[1 - (0.1)^6]}{1 - 0.1} \\[1em] = \dfrac{0.1(1 - 0.000001)}{0.9} \\[1em] = \dfrac{0.1(0.999999)}{0.9} \\[1em] = \dfrac{0.1(0.999999)}{0.9} \\[1em] = \dfrac{1}{9} \times 0.999999 \\[1em] = 0.111111.$

Hence, S₆ = 0.111111.

Question 5

Find the sum of first 6 terms of the G.P. 1, $-\dfrac{1}{3}, \dfrac{1}{3^{2}}, -\dfrac{1}{3^{3}}$ , .......

Answer

Given,

a = 1

r = $\dfrac{\dfrac{-1}{3}}{1} = -\dfrac{1}{3}$

n = 6

We know that,

The sum of the first n terms of a G.P. is given by:

$S_n = \dfrac{a(1 - r^n)}{1 - r}$ [For r < 1]

Substituting values we get :

$\Rightarrow S_6 = \dfrac{1\Big[1 - \Big(\dfrac{-1}{3}\Big)^6\Big]}{1 - \Big(\dfrac{-1}{3}\Big) } \\[1em] = \dfrac{\Big(1 - \dfrac{1}{3^6}\Big)}{1 + \dfrac{1}{3}} \\[1em] = \dfrac{\Big(1 - \dfrac{1}{729}\Big)}{\dfrac{3 + 1}{3}} \\[1em] = \dfrac{\Big(\dfrac{729 - 1}{729}\Big)}{\dfrac{4}{3}} \\[1em] = \dfrac{\dfrac{728}{729}}{\dfrac{4}{3}} \\[1em] = \dfrac{728}{729}\times \dfrac{3}{4} \\[1em] = \dfrac{182}{243}.$

Hence, S₆ = $\dfrac{182}{243}$ .

Question 6

The first term of a G.P. is 27 and its 8th term is $\dfrac{1}{81}$ . Find the sum of first seven terms of the G.P.

Answer

Given,

a = 27

$\Rightarrow T_8 = \dfrac{1}{81} \\[1em] \Rightarrow ar^{(8-1)} = \dfrac{1}{81} \\[1em] \Rightarrow ar^{7} = \dfrac{1}{81} \\[1em] \Rightarrow (27)r^{7} = \dfrac{1}{81} \\[1em] \Rightarrow r^{7} = \dfrac{1}{81 \times 27} \\[1em] \Rightarrow r^{7} = \dfrac{1}{3^4 \times 3^3} \\[1em] \Rightarrow r^{7} = \dfrac{1}{3^{4 + 3}} \\[1em] \Rightarrow r^{7} = \dfrac{1}{3^{7}} \\[1em] \Rightarrow r^{7} = \Big(\dfrac{1}{3}\Big)^7 \\[1em] \Rightarrow r = \dfrac{1}{3}.$

We know that,

The sum of the first n terms of a G.P. is given by:

$S_n = \dfrac{a(1 - r^n)}{1 - r}$ [For r < 1]

Substituting values we get :

$\Rightarrow S_7 = \dfrac{27\Big[1 - \Big(\dfrac{1}{3}\Big)^7\Big]}{1 - \Big(\dfrac{1}{3}\Big)} \\[1em] = \dfrac{27\Big[1 - \Big(\dfrac{1}{2187}\Big)\Big]}{\Big(\dfrac{3 - 1}{3}\Big)} \\[1em] = \dfrac{27\Big(\dfrac{2187 - 1}{2187}\Big)}{\Big(\dfrac{3 - 1}{3}\Big)} \\[1em] = \dfrac{27\Big(\dfrac{2186}{2187}\Big)}{\Big(\dfrac{2}{3}\Big)} \\[1em] = \dfrac{27 \times 2186 \times 3}{2 \times 2187} \\[1em] = \dfrac{1093}{27}.$

Hence, S₇ = $\dfrac{1093}{27}$ .

Question 7

The 4th and the 7th terms of a G.P. are $\dfrac{1}{27}$ and $\dfrac{1}{729}$ respectively. Find the sum of first 6 terms of the G.P.

Answer

Given,

⇒ T₄ = $\dfrac{1}{27}$

⇒ ar³ = $\dfrac{1}{27}$ .....(1)

Given,

⇒ T₇ = $\dfrac{1}{729}$

⇒ ar⁶ = $\dfrac{1}{729}$ ..........(2)

Dividing Equation (2) by Equation (1) :

$\Rightarrow \dfrac{ar^6}{ar^3} = \dfrac{\dfrac{1}{729}}{\dfrac{1}{27}} \\[1em] \Rightarrow r^{6 - 3} = \dfrac{1}{729} \times 27 \\[1em] \Rightarrow r^{3} = \dfrac{1}{27} \\[1em] \Rightarrow r = \sqrt[3]{\dfrac{1}{27}} \\[1em] \Rightarrow r = \dfrac{1}{3}.$

Substituting, $r = \dfrac{1}{3}$ in equation 1 :

$\Rightarrow ar^3 = \dfrac{1}{27} \\[1em] \Rightarrow a\Big(\dfrac{1}{3}\Big)^3 = \dfrac{1}{27} \\[1em] \Rightarrow a\Big(\dfrac{1}{27}\Big) = \dfrac{1}{27} \\[1em] \Rightarrow a = \dfrac{1}{27} \times 27 \\[1em] \Rightarrow a = 1.$

We know that,

The sum of the first n terms of a G.P. is given by:

$S_n = \dfrac{a(1 - r^n)}{1 - r}$ [For r < 1]

Substituting values we get :

$\Rightarrow S_6 = \dfrac{1\Big[1 - \Big(\dfrac{1}{3}\Big)^6\Big]}{1 - \Big(\dfrac{1}{3}\Big)} \\[1em] = \dfrac{\Big(1 - \dfrac{1}{729}\Big)}{\dfrac{3 - 1}{3}} \\[1em] = \dfrac{\Big(\dfrac{729 - 1}{729}\Big)}{\Big(\dfrac{3 - 1}{3}\Big)} \\[1em] = \dfrac{\Big(\dfrac{728}{729}\Big)}{\Big(\dfrac{2}{3}\Big)} \\[1em] = \dfrac{728}{729} \times \dfrac{3}{2} \\[1em] = \dfrac{364}{243}.$

Hence, S₆ = $\dfrac{364}{243}$ .

Question 8

How many terms of the G.P. $\dfrac{2}{9}$ , $-\dfrac{1}{3}$ , $\dfrac{1}{2}$ , ……… must be taken to make the sum equal to $\dfrac{55}{72}$ ?

Answer

In the given G.P.,

a = $\dfrac{2}{9}$

r = $\dfrac{\dfrac{-1}{3}}{\dfrac{2}{9}} = \dfrac{-9}{2 \times 3} = \dfrac{-3}{2}$ .

Let sum of n terms be equal to $\dfrac{55}{72}$ .

S_n = $\dfrac{55}{72}$ .

We know that,

The sum of the first n terms of a G.P. is given by :

$S_n = \dfrac{a(1 - r^n)}{1 - r}$ [For r < 1]

Substituting values we get :

$\Rightarrow \dfrac{55}{72} = \dfrac{\dfrac{2}{9}\Big[1 - \Big(-\dfrac{3}{2}\Big)^n\Big]}{1 - \Big(-\dfrac{3}{2}\Big)} \\[1em] \Rightarrow \dfrac{55}{72} = \dfrac{\dfrac{2}{9}\Big[1 - \Big(-\dfrac{3}{2}\Big)^n\Big]}{\Big(1 + \dfrac{3}{2}\Big)} \\[1em] \Rightarrow \dfrac{55}{72} = \dfrac{\dfrac{2}{9}\Big[1 - \Big(-\dfrac{3}{2}\Big)^n\Big]}{\Big(\dfrac{5}{2}\Big)} \\[1em] \Rightarrow \dfrac{55}{72} \times \Big(\dfrac{5}{2}\Big)= \dfrac{2}{9}\Big[1 - \Big(-\dfrac{3}{2}\Big)^n\Big] \\[1em] \Rightarrow \dfrac{275}{144} = \dfrac{2}{9}\Big[1 - \Big(-\dfrac{3}{2}\Big)^n\Big] \\[1em] \Rightarrow \dfrac{275}{144} \times \dfrac{9}{2} = \Big[1 - \Big(-\dfrac{3}{2}\Big)^n\Big] \\[1em] \Rightarrow \dfrac{275}{32} = 1 - \Big(-\dfrac{3}{2}\Big)^n \\[1em] \Rightarrow \Big(-\dfrac{3}{2}\Big)^n = 1 - \dfrac{275}{32} \\[1em] \Rightarrow \Big(-\dfrac{3}{2}\Big)^n = \dfrac{32 - 275}{32} \\[1em] \Rightarrow \Big(-\dfrac{3}{2}\Big)^n = \dfrac{-243}{32} \\[1em] \Rightarrow \Big(-\dfrac{3}{2}\Big)^n = \Big(-\dfrac{3}{2}\Big)^5 \\[1em] \Rightarrow n = 5.$

Hence, n = 5.

Question 9

In a G.P. the ratio of the sum of first 3 terms is to that of first 6 terms is 125 : 152. Find the common ratio.

Answer

Let first term of G.P. be a and common ratio be r.

Given,

$\Rightarrow \dfrac{S_3}{S_6} = \dfrac{125}{152} \\[1em] \Rightarrow \dfrac{\dfrac{a(1 - r^3)}{1 - r}}{\dfrac{a(1 - r^6)}{1 - r}} = \dfrac{125}{152} \\[1em] \Rightarrow \dfrac{(1 - r^3)}{(1 - r^6)} = \dfrac{125}{152} \\[1em] \Rightarrow \dfrac{(1 - r^3)}{(1)^2 - (r^3)^2} = \dfrac{125}{152} \\[1em] \Rightarrow \dfrac{(1 - r^3)}{(1 + r^3)(1 - r^3)} = \dfrac{125}{152} \\[1em] \Rightarrow \dfrac{1}{(1 + r^3)} = \dfrac{125}{152} \\[1em] \Rightarrow 1 + r^3 = \dfrac{152}{125} \\[1em] \Rightarrow r^3 = \dfrac{152}{125} - 1 \\[1em] \Rightarrow r^3 = \dfrac{152 - 125}{125} \\[1em] \Rightarrow r^3 = \dfrac{27}{125} \\[1em] \Rightarrow r = \sqrt[3]{\dfrac{27}{125}} \\[1em] \Rightarrow r = \dfrac{3}{5}.$

Hence, r = $\dfrac{3}{5}$ .

Question 10

A manufacturer reckons that the value of a machine which costs him ₹ 31,250 depreciates each year by 20%. Find its value after 2 years.

Answer

Given,

The initial cost of machine is ₹ 31,250. Therefore,

a = ₹ 31,250

Depreciation Rate : 20% per year

If 20% is lost, the percentage of the value retained is: 100% - 20% = 80%.

r = $\dfrac{80}{100}$ [constant factor by which the value of the machine is multiplied each year to get the next year's value.]

Since, value after 2 years is the value in the beginning of third year, thus n = 3.

We know that,

$\Rightarrow T_n = ar^{(n-1)} \\[1em] \Rightarrow T_3 = ar^{3-1} \\[1em] = 31250 \Big(\dfrac{80}{100}\Big)^2 \\[1em] = 31250 (0.8)^2 \\[1em] = 31250 (0.64) \\[1em] = 20,000.$

Hence, value of machine after 2 years = ₹20,000.

Question 11

Find the sum of the following to n terms: 7 + 77 + 777 + 7777 + ………

Answer

$S_n = 7 + 77 + 777 + 7777 + ……… \text{ upto n terms} \\[1em] = 7(1 + 11 + 111 +.....) \text{ upto n terms} \\[1em] = \dfrac{7}{9} (9 + 99 + 999 +.....) \text{ upto n terms} \\[1em] = \dfrac{7}{9}[(10 - 1) + (10^2 - 1) + (10^3 - 1)+.......+(10^n - 1)] \\[1em] = \dfrac{7}{9}[(10 + 10^2 + 10^3+.....+10^n) - (1 + 1 + 1+.....\text{n times})] \\[1em] = \dfrac{7}{9}[(10 + 10^2 + 10^3+.....+10^n) - n] \text{ .......(1)}$

Now,

Calculating the sum of 10 + 10² + 10³ + ......... + 10ⁿ.

a = 10

r = $\dfrac{10^2}{10}$ = 10

We know that,

The sum of the first n terms of a G.P. is given by:

$S_n = \dfrac{a(r^n - 1)}{r - 1}$ [For r > 1]

$\Rightarrow S_n = \dfrac{10(10^n - 1)}{10 - 1} \\[1em] = \dfrac{10(10^n - 1)}{9}.$

Substitute and Simplify, the above value in equation (1), we get :

$S_n = \dfrac{7}{9}\Big[\dfrac{10(10^n - 1)}{9} - n\Big] \\[1em] = \dfrac{7}{81}\Big[10 \times 10^n - 10 - 9n\Big] \\[1em] = \dfrac{7}{81}\Big[10^{n + 1} - 9n - 10\Big]$

Hence, S_n = $\dfrac{7}{81}\Big[10^{n + 1} - 9n - 10\Big]$ .

Question 12

Find the sum of n terms of series whose mth term is 2^m + 2m

Answer

Given,

mth term = 2^m + 2m

1st term = 2¹ + 2 × 1

2nd term = 2² + 2 × 2

nth term = 2ⁿ + 2 × n

$S_n = [(2^1 + 2 \times 1) + (2^2 + 2 \times 2) +(2^3 + 2 \times 3)+........ + (2^n + 2 \times n)] \\[1em] = [2^1 + 2^2 + 2^3 + ..... 2^n + (2 \times 1 + 2 \times 2 + 2 \times 3 + ..... + 2 \times n)] \\[1em] = (2^1 + 2^2 + 2^3 + ..... + 2^n) + 2(1 + 2 + 3 + ..... + n)......(1)$

Calculating :

2¹ + 2² + ........ + 2ⁿ

The above is an G.P. with a = 2 and r = 2.

By using formula,

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

Substitute values, we get:

S_n = $2 \cdot \dfrac{2^n - 1}{2 - 1}$

= 2(2ⁿ - 1)

Calculating :

(1 + 2 + 3 + ..... + n)

The above is an A.P. with a = 1 and d = 1.

By using formula,

S_n = $\dfrac{n}{2}$ [2a + (n - 1)d]

Substitute values, we get:

S_n = $\dfrac{n}{2}$ [2(1) + (n - 1)1]

= $\dfrac{n}{2}$ [2 + n - 1]

= $\dfrac{n}{2}$ (n + 1).

Substitute values in (1) we get,

S_n = 2(2ⁿ - 1) + $2 \times \dfrac{n}{2}$ (n + 1)

= 2(2ⁿ - 1) + n(n + 1)

Hence, S_n = 2(2ⁿ - 1) + n(n + 1).

Multiple Choice Questions

Question 1

The nth term of a G.P. whose first term is a and common ratio is r, is given by :

T_n = ar^{n + 1}
T_n = ar^{1 - n}
T_n = ar^{n - 1}
T_n = $\dfrac{a}{r^{,n-1}}$

Answer

The nth term of a G.P. whose first term is a and common ratio is r is given by : ar^{n - 1}.

Hence, option 3 is the correct option.

Question 2

If the first term, common ratio and the last term of a G.P. are a, r and l respectively, then the nth term from the end of the G.P. is given by :

lr^{n - 1}
lr^{1 - n}
$\dfrac{l}{r^{1 - n}}$
$\dfrac{l}{r^{n + 1}}$

Answer

When the G.P. is reversed, the first term becomes l and the new common ratio is the reciprocal of the original common ratio i.e. $\dfrac{1}{r}$ .

$\therefore T_n = l \times \Big(\dfrac{1}{r}\Big)^{n - 1} \\[1em] = \dfrac{l}{r^{n - 1}} \\[1em] = lr^{-(n - 1)} \\[1em] = lr^{1 - n}.$

Hence, option 2 is the correct option.

Question 3

The product of n terms of a G.P. with first term a and common ratio r, when r = 1, is :

$\dfrac{a(r^n-1)}{r-1}$
na
arⁿ
aⁿ

Answer

Given,

r = 1, then every term is same : a, a, a......a

Product = a × a × a × ...... upto n terms = aⁿ.

Hence, option 4 is the correct option.

Question 4

The sum of n terms of a G.P. with first term a and common ratio r, when r = 1, is given by:

$S_n = \dfrac{a(1-r^n)}{1-r}$
$S_n = \dfrac{a(r^n-1)}{r-1}$
n²a
na

Answer

Given,

r = 1, then all terms are same:

a, a, a, ......a

S_n = a + a + a + a...... + upto n terms

= n × a.

Hence, option 4 is the correct option.

Question 5

The sum of n terms of a G.P. with first term a and common ratio r, when r > 1, is :

$\dfrac{ar^n-1}{r-1}$
$\dfrac{1-ar^n}{1-r}$
$\dfrac{a(r^n-1)}{r-1}$
$\dfrac{a(1-r^n)}{1-r}$

Answer

Formula for sum of n terms of a G.P.

For a geometric progression with first term = a and common ratio = r > 1,

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

Hence, option 3 is the correct option.

Question 6

The sum of n terms of a G.P. with first term a and common ratio r, when r < 1, is :

$\dfrac{ar^n-1}{r-1}$
$\dfrac{1-ar^n}{1-r}$
$\dfrac{a(r^n-1)}{r-1}$
$\dfrac{a(1-r^n)}{1-r}$

Answer

Formula for sum of n terms of a G.P.

For a geometric progression with first term = a and common ratio (r) < 1,

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

Hence, option 4 is the correct option.

Question 7

The general term of the G.P. $\dfrac{1}{4}, -\dfrac{1}{2}, 1, -2, 4, \dots$ is :

(-1)^{(n - 1)} × 2^{(n - 3)}
(-1)^{(n - 1)} × 2^{(n - 2)}
(-1)^{(n - 1)} × (-2)^{(n - 1)}
(-1)^{(n - 1)} × (-2)^{(n - 3)}

Answer

We know that,

nth term of a G.P. is given by,

T_n = ar^{n - 1}

In the given G.P.,

a = $\dfrac{1}{4}$

r = $\dfrac{-\dfrac{1}{2}}{\dfrac{1}{4}}$ = -2.

⇒ T_n = $\dfrac{1}{4}$ (-2)^{n - 1}

= $\dfrac{1}{2^2}$ .(-2)^{n - 1}

= 2^-2.(-1)^{n - 1}.(2)^{n - 1}

= (-1)^{n - 1}.(2)^{-2 + n - 1}

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

Hence, option 1 is the correct option.

Question 8

The 12th term of the G.P. 2, 4, 8, 16, ....... is :

1024
2048
4096
8192

Answer

We know that,

nth term of a G.P. is given by,

T_n = ar^{n - 1}

In the given A.P.,

a = 2

r = $\dfrac{4}{2}$ = 2

n = 12

⇒ T₁₂ = 2.(2)^{12 - 1}

= 2(2)¹¹

= (2)^{11 + 1}

= (2)¹²

= 4096.

Hence, option 3 is the correct option.

Question 9

Which term of the G.P. $\sqrt{3}, 3\sqrt{3}, 9\sqrt{3}, \dots$ is $729\sqrt{3}$ ?

Answer

In the given A.P.,

a = $\sqrt3$

r = $\dfrac{3\sqrt3}{\sqrt3}$ = 3

Let nth term of G.P. be $729\sqrt{3}$ .

⇒ T_n = $729\sqrt3$

⇒ $\sqrt{3} \times 3^{n - 1} = 729\sqrt{3}$

⇒ (3)^{n - 1} = 729

⇒ (3)^{n - 1} = 3⁶

⇒ n - 1 = 6

⇒ n = 6 + 1

⇒ n = 7.

Hence, option 1 is the correct option.

Question 10

If a₁, a₂, a₃, ......., a_n is a G.P. having common ratio r and k is a natural number such that 3 < k < n, then r is equal to :

$\dfrac{a_k}{a_{1}}$
$\dfrac{a_1}{a_2}$
$\dfrac{a_k}{a_{n-3}}$
$\dfrac{a_{k-1}}{a_{k-2}}$

Answer

We know that,

T_n = ar^{n - 1},

In the G.P.,

a₁, a₂, a₃, ......., a_n

a₁ is the first term and r is the common ratio.

$\Rightarrow \dfrac{a_{k - 1}}{a_{k - 2}} \\[1em] = \dfrac{a_1r^{(k - 1) - 1}}{a_1r^{(k - 2) - 1}} \\[1em] = \dfrac{a_1r^{k - 2}}{a_1r^{k - 3}} \\[1em] = r^{k - 2 - (k - 3)} \\[1em] = r^{k - 2 - k + 3} \\[1em] = r^{k - k + 3 - 2} \\[1em] = r.$

Hence, option 4 is the correct option.

Question 11

The common ratio of the G.P. $-\dfrac{3}{4}, \dfrac{1}{2}, -\dfrac{1}{3}, \dfrac{2}{9}, \dots$ is :

$-\dfrac{4}{3}$
$-\dfrac{2}{3}$
$\dfrac{2}{3}$
$-\dfrac{3}{8}$

Answer

r = $\dfrac{\dfrac{1}{2}}{-\dfrac{3}{4}}$

= $\dfrac{1}{2} \times -\dfrac{4}{3}$

= $-\dfrac{2}{3}$ .

Hence, option 2 is the correct option.

Question 12

The common ratio of the G.P. $\dfrac{1}{a^3x^3},\ ax,\ a^5x^5,\ \dots$ is :

$\dfrac{1}{a^2x^2}$
$\dfrac{1}{a^4x^4}$
a²x²
a⁴x⁴

Answer

$r = \dfrac{ax}{\dfrac{1}{a^3x^3}} \\[1em] = ax \times a^3x^3 = a^4x^4.$

Hence, option 4 is the correct option.

Question 13

The common ratio of the G.P. 0.15, 0.015, 0.0015, ...... is :

0.1
0.01
1
0.001

Answer

r = $\dfrac{0.015}{0.15}$

= 0.1

Hence, option 1 is the correct option.

Question 14

The 8th term of the G.P. 1, 3, 9, 27, .......... is :

729
2187
6561
none of these

Answer

In the given G.P.,

a = 1

r = $\dfrac{3}{1}$ = 3

n = 8

We know that,

nth term of a G.P. is given by,

T_n = ar^{n - 1}

⇒ T₈ = 1.(3)^{8 - 1}

= (3)⁷

= 2187.

Hence, option 2 is the correct option.

Question 15

The nth term of the G.P. x³, x⁵, x⁷, ........ is :

x^{(2n - 1)}
x^{(2n + 3)}
x^{(2n + 1)}
x^{3n + 2}

Answer

In the given G.P.,

a = x³

r = $\dfrac{x^5}{x^3}$ = x^{5 - 3} = x²

We know that,

nth term of a G.P. is given by,

T_n = ar^{n - 1}

⇒ T_n = x³.(x²)^{n - 1}

= x³.(x^{2n - 2})

= x^{2n - 2 + 3}

= x^{2n + 1}.

Hence, option 3 is the correct option.

Question 16

The 10th term of the G.P. 1, −a, a², −a³, ........ is :

a⁹
−a¹⁰
−a¹¹
−a⁹

Answer

In the given G.P.,

a = 1

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

n = 10

We know that,

nth term of a G.P. is given by,

⇒ T_n = ar^{n - 1}

⇒ T₁₀ = 1.(-a)^{10 - 1}

= (-a)⁹.

Hence, option 4 is the correct option.

Question 17

The first term and the common ratio of the G.P. 3, $\dfrac{3}{2}, \dfrac{3}{4}$ , .......... are respectively :

3 and 2
3 and $\dfrac{1}{2}$
3 and $\dfrac{3}{2}$
$\dfrac{3}{2}$ and $\dfrac{1}{2}$

Answer

In the given G.P.,

a = 3

r = $\dfrac{\dfrac{3}{2}}{3} = \dfrac{3}{2} \times \dfrac{1}{3} = \dfrac{1}{2}$ .

Hence, option 2 is the correct option.

Question 18

Which term of the G.P. 2, 8, 32, 128, ........ is 131072?

9th
10th
8th
12th

Answer

In the given G.P.,

a = 2

r = $\dfrac{8}{2}$ = 4

We know that,

T_n = ar^{n - 1}

Let nth term be 131072.

⇒ T_n = 131072

⇒ 2.(4)^{n - 1} = 131072

⇒ (2²)^{n - 1} = $\dfrac{131072}{2}$

⇒ 2^{2n - 2} = 65536

⇒ 2^{2n - 2} = 2¹⁶

Equate exponents:

⇒ 2n - 2 = 16

⇒ 2n = 16 + 2

⇒ 2n = 18

⇒ n = $\dfrac{18}{2}$ = 9.

Hence, option 1 is the correct option.

Question 19

If the 5th term of a G.P. is 2, then the product of its first nine terms is :

256
1024
512
2048

Answer

Let first term and common ratio of G.P. be a and r respectively.

Given,

5th term of a G.P. is 2.

⇒ T₅ = 2

⇒ ar^{5 - 1} = 2

⇒ ar⁴ = 2 .....(1)

Product of first 9 terms = a × ar¹ × ar² × ...... × ar⁸

⇒ a⁹.r^{0 + 1 + 2 + .... + 8}

⇒ a⁹.r³⁶

⇒ (ar⁴)⁹ .....(2)

Substituting value of ar⁴ from equation (1) in (2), we get :

⇒ (2)⁹

⇒ 512.

Hence, option 3 is the correct option.

Question 20

If the (p + q)th and (p − q)th term of a G.P. are m and n respectively, then its pth term is :

mn
$\sqrt{mn}$
(mn)²
$\dfrac{m}{n}$

Answer

Given,

(p + q)th term = m

(p - q)th term = n

We know that,

nth term of a G.P. is given by,

T_n = ar^{n - 1}

⇒ T_{p + q} = m

⇒ ar^{p + q - 1} = m ........(1)

⇒ T_{p - q} = n

⇒ ar^{p - q - 1} = n ........(2)

⇒ T_p = ar_{p - 1}

Multiplying equation (1) and (2) :

⇒ ar^{p + q - 1} × ar^{p - q - 1} = mn

⇒ a².r^{p + q - 1 + p - q - 1} = mn

⇒ a².r^{2p - 2} = mn

Taking square root on both sides:

⇒ $\sqrt{a^2r^{2p - 2}} = \sqrt{mn}$

⇒ ar^{p - 1} = $\sqrt{mn}$

Thus, pth term = $\sqrt{mn}$ .

Hence, option 2 is the correct option.

Question 21

If 2nd, 3rd and 6th terms of an A.P. are the three consecutive terms of a G.P., then the common ratio of the G.P. is :

2
3
$\dfrac{1}{2}$
$\dfrac{1}{3}$

Answer

Let first term of A.P. be a and common difference be d.

2nd Term : a + d

3rd Term : a + 2d

6th Term : a + 5d

These three terms form three consecutive terms of a G.P.

Thus, a + d, a + 2d, a + 5d, ........ is the G.P.

In G.P., ratio between consecutive terms are equal.

$\Rightarrow \dfrac{a + 5d}{a + 2d} = \dfrac{a + 2d}{a + d}$

⇒ (a + 2d)² = (a + d)(a + 5d)

⇒ a² + 4ad + 4d² = a² + 5ad + ad + 5d²

⇒ a² + 4ad + 4d² = a² + 6ad + 5d²

⇒ 0 = a² - a² + 6ad - 4ad + 5d² - 4d²

⇒ 0 = 2ad + d²

⇒ d(2a + d) = 0

⇒ d = 0 or 2a + d = 0

d cannot be equal to zero as then common ratio will be equal to 1, also not in options.

⇒ 2a + d = 0

⇒ d = -2a

Substitute d = −2a :

a + d = a - 2a = -a

a + 2d = a + 2(-2a) = -3a

a + 5d = a + 5(-2a) = -9a

r = $\dfrac{-3a}{-a}$ = 3.

Hence, option 2 is the correct option.

Question 22

The 6th term from the end of the G.P. 8, 4, 2, ......, $\dfrac{1}{1024}$ is :

$\dfrac{1}{32}$
$\dfrac{1}{16}$
$\dfrac{1}{64}$
$\dfrac{1}{128}$

Answer

G.P. : 8, 4, 2, ......, $\dfrac{1}{1024}$ .

a = 8

r = $\dfrac{4}{8} = \dfrac{1}{2}$

l = $\dfrac{1}{1024}$ .

We know that,

${\text{nth term from the end}} = \dfrac{l}{r^{n - 1}}$

Substitute values we get:

$\Rightarrow {\text{6th term from the end}} = \dfrac{l}{r^{6 - 1}} \\[1em] = \dfrac{\dfrac{1}{1024}}{\Big(\dfrac{1}{2}\Big)^{5}} \\[1em] = \dfrac{1}{1024} \times \dfrac{32}{1}\\[1em] = \dfrac{1}{32}.$

Hence, the 6th term from the end is $\dfrac{1}{32}$ .

Hence, option 1 is the correct option.

Question 23

The 4th term from the end of the G.P. $\dfrac{2}{27}, \dfrac{2}{9}, \dfrac{2}{3}, ......, 162$ is :

18
2
6
$\dfrac{2}{3}$

Answer

G.P. : $\dfrac{2}{27}, \dfrac{2}{9}, \dfrac{2}{3}, ......, 162$ .

a = $\dfrac{2}{27}$

r = $\dfrac{\dfrac{2}{9}}{\dfrac{2}{27}} = \dfrac{2}{9} \times \dfrac{27}{2}$ = 3

l = 162.

We know that,

${\text{nth term from the end}} = \dfrac{l}{r^{n - 1}}$

Substitute values we get:

$\Rightarrow {\text{4th term from the end}} = \dfrac{l}{r^{4 - 1}} \\[1em] = \dfrac{162}{3^{3}} \\[1em] = \dfrac{162}{27} \\[1em] = 6.$

Hence, the 6th term from the end is 6.

Hence, option 3 is the correct option.

Question 24

The product of first three terms of a G.P. is −1 and the common ratio is $-\dfrac{3}{4}$ . The sum of these three terms is :

$\dfrac{12}{13}$
$\dfrac{11}{13}$
$\dfrac{11}{12}$
$\dfrac{13}{12}$

Answer

Let the first three terms of the G.P. be $\dfrac{a}{r}$ , a, ar.

The product of the three terms is given as -1

⇒ $\dfrac{a}{r}$ × a × ar = -1

⇒ a³ = -1

⇒ a = $\sqrt[3]{-1}$

⇒ a = -1.

Substitute a = -1 and r = $\dfrac{-3}{4}$ , we get :

⇒ $\dfrac{a}{r} = \dfrac{-1}{\dfrac{-3}{4}} = \dfrac{4}{3}$ ,

⇒ ar = $(-1) \times \Big(-\dfrac{3}{4}\Big) = \dfrac{3}{4}$ .

The sum of the three terms is :

$\Rightarrow \dfrac{4}{3} + (-1) + \dfrac{3}{4} \\[1em] = \dfrac{16 - 12 + 9}{12} \\[1em] = \dfrac{25 - 12}{12} \\[1em] = \dfrac{13}{12}.$

Hence, option 4 is the correct option.

Question 25

For what values of x are the numbers $-\dfrac{2}{7}, x, -\dfrac{7}{2}$ in G.P.?

0, 1
0, −1
−1, 1
−2, 2

Answer

We know that,

The numbers are in G.P., if the ratio of consecutive terms are equal.

$\Rightarrow \dfrac{x}{-\dfrac{2}{7}} = \dfrac{-\dfrac{7}{2}}{x} \\[1em] \Rightarrow x^2 = \dfrac{-7}{2} \times \dfrac{-2}{7} \\[1em] \Rightarrow x^2 = \dfrac{14}{14} \\[1em] \Rightarrow x^2 = 1 \\[1em] \Rightarrow x = 1 \text { or } x = -1.$

Hence, option 3 is the correct option.

Question 26

How many terms of the G.P. 1, 4, 16, 64, ........ will make the sum 5461?

Answer

In the given G.P.,

a = 1

r = $\dfrac{4}{1}$ = 4

Formula for sum of n terms of a G.P.

$S_n = \dfrac{a(r^n-1)}{r-1}$ [r > 1]

Let the sum of n terms of the G.P. = 5461.

$\Rightarrow S_n = 5461 \\[1em] \Rightarrow 5461 = \dfrac{1(4^n - 1)}{(4 - 1)} \\[1em] \Rightarrow 5461 = \dfrac{(4^n - 1)}{3} \\[1em] \Rightarrow 5461 \times 3 = 4^n - 1 \\[1em] \Rightarrow 16383 = 4^n - 1 \\[1em] \Rightarrow 16383 + 1 = (2^2)^n \\[1em] \Rightarrow 16384 = 2^{2n} \\[1em] \Rightarrow 2^{14} = 2^{2n} \\[1em] \Rightarrow 2n = 14 \\[1em] \Rightarrow n = \dfrac{14}{2} \\[1em] \Rightarrow n = 7.$

Hence, option 3 is the correct option.

Question 27

The sum of 7 terms of the G.P. 3, 6, 12, .......... is :

Answer

In the given G.P.,

a = 3

r = $\dfrac{6}{3}$ = 2

n = 7

Formula for sum of n terms of a G.P.

$S_n = \dfrac{a(r^n-1)}{r-1}$ [r > 1]

Substituting values we get :

$\Rightarrow S_7 = \dfrac{3(2^7 - 1)}{2-1} \\[1em] = \dfrac{3(2^7 - 1)}{1} \\[1em] = 3(128 - 1) \\[1em] = 3 \times (127) \\[1em] = 381.$

Hence, option 3 is the correct option.

Question 28

The sum of the first two terms of a G.P. is −4 and the fifth term is 4 times the third term. Then, the first term of the G.P. is :

$-\dfrac{3}{4}$ or 4
$\dfrac{3}{4}$ or $\dfrac{1}{4}$
$\dfrac{4}{3}$ or $\dfrac{1}{4}$
$-\dfrac{4}{3}$ or 4

Answer

Let the first term be a and the common ratio be r.

Given,

Sum of the first two terms is -4

⇒ a + ar = -4

⇒ a(1 + r) = -4 .....(1)

Given,

The fifth term is 4 times the third term.

⇒ ar^{5 - 1} = 4ar^3-1

⇒ ar⁴ = 4ar²

⇒ r² = 4

⇒ r = $\sqrt{4}$

⇒ r = 2 or r = -2

Substituting r = 2 into equation 1 :

⇒ a(1 + 2) = -4

⇒ 3a = -4

⇒ a = $-\dfrac{4}{3}$ .

Substituting r = -2 into equation 1:

⇒ a[1 + (-2)] = -4

⇒ a(1 - 2) = -4

⇒ a(-1) = -4

⇒ a = 4.

Hence, option 4 is the correct option.

Question 29

If x, y and z are in G.P., then the relation between x, y and z can be :

y = x + z
y = xz
2y = x + z
$y = \sqrt{xz}$

Answer

Given,

x, y and z are in G.P.

Ratio between consecutive terms are equal in a G.P.

$\Rightarrow \dfrac{y}{x} = \dfrac{z}{y} \\[1em] \Rightarrow y^2 = xz \\[1em] \Rightarrow y = \sqrt{xz}.$

Hence, option 4 is the correct option.

Question 30

The product of first five terms of a G.P. with first term a and common ratio r > 1 is equal to :

$\dfrac{a(r^5 - 1)}{r - 1}$
ar⁴
a⁵r¹⁰
$\dfrac{a(r^5 - 1)}{r}$

Answer

Given,

First term = a

Common ratio = r

We know that,

T_n = ar^{(n - 1)}

The product of first five terms of a G.P.

P = a × ar × ar² × ar³ × ar⁴

= a⁵ × r^0+1+2+3+4

= a⁵.r¹⁰

Hence, option 3 is the correct option.

Question 31

If 5th, 8th and 11th terms of a G.P. are x, y and z respectively, then which one of the following is correct?

y² = x²z²
y² = x² + z²
y² = xz
xyz = 1

Answer

Let first term of G.P. be a and common ratio be r.

Given,

5th term = x

x = ar^{5 - 1}

x = ar⁴

8th term = y

y = ar^{8 - 1}

y = ar⁷

11th term = z

z = ar^{11 - 1}

z = ar¹⁰

Substituting value of y in L.H.S. of y² = xz

⇒ y²

⇒ (ar⁷)²

⇒ (a²r¹⁴)

Substituting value of x and z in R.H.S. of y² = xz

⇒ xz

⇒ ar⁴ × ar¹⁰

⇒ a²r¹⁴.

Since, R.H.S. = L.H.S.

Hence proved, that y² = xz.

Hence, option 3 is the correct option.

Question 32

Consider the G.P. a, ar, ar², ........, l. The kth term from the end is :

$\dfrac{l}{r^{k-1}}$
$\dfrac{l}{k}$
$\dfrac{al}{r^k}$
$\dfrac{l}{ar^{k-1}}$

Answer

Given,

G.P. a, ar, ar², ........, l.

We know that,

$\text{n from end} = \dfrac{l}{r^{n - 1}}$

$\Rightarrow \text{kth from end} = \dfrac{l}{r^{k - 1}}$

Hence, option 1 is the correct option.

Question 33

If a and l are respectively the first and the last terms of a G.P. having common ratio r > 1, then the sum to n terms of the G.P. is given by :

lar^{n - 1}
ar^ln
$\dfrac{lr - a}{r - 1}$
$\dfrac{lr^n - a}{lr - 1}$

Answer

The standard formula for the sum of the first n terms of a G.P. is :

$\Rightarrow S_n = \dfrac{a(r^n - 1)}{r - 1} \\[1em] \Rightarrow S_n = \dfrac{ar^n - a}{r - 1} \\[1em] \Rightarrow S_n = \dfrac{ar^{n - 1} \times r - a}{ r - 1} \text{ .....(1)}$

The last term of the G.P. is the nth term.

l = ar^{n - 1} ......(2)

Substituting value of ar^{n - 1} from equation (2) in (1), we get :

$S_n = \dfrac{lr - a}{r - 1}$

Hence, option 3 is the correct option.

Question 34

The product of n terms of a G.P. with first term a and common ratio r, r > 1 is :

$\dfrac{a^n(r^{n} - 1)}{r - 1}$
aⁿrⁿ
$a^{n} r^{\dfrac{n(n+1)}{2}}$
$a^{n} r^{\dfrac{n(n-1)}{2}}$

Answer

Given,

G.P. : a, ar, ar², ar³,.....,ar^{n - 1}

Product = a × ar × ar² × ar³×.....×ar^{n - 1}

= aⁿ × r^{0 + 1 + 2 + 3 + .... + (n-1)}

= $a^nr^\dfrac{n(n-1)}{2}$ [As sum of first n - 1 natural numbers = $\dfrac{n(n - 1)}{2}$ ]

Hence, option 4 is the correct option.

Case Study Based Questions

Question 1

Case Study

The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, then

Based on the above given information, answer the following questions:

The number of bacteria that would be present at the end of 4th hour is :
(a) 240
(b) 480
(c) 960
(d) 450
The number of bacteria that would be present at the end of 10th hour is :
(a) 30720
(b) 15360
(c) 61440
(d) 27540
The number of bacteria that would be present at the end of nth hour is :
(a) 30 × 2^{(n - 1)}
(b) 30 × 2^{(n + 1)}
(c) 30 × 2ⁿ
(d) none of these
What would be the sum of the first 5 terms of the G.P. that is formed by the number of bacteria after the completion of each hour?
(a) 450
(b) 930
(c) 1890
(d) none of these
What would be the sum of the first n terms of the G.P. that is formed by the number of bacteria after the completion of each hour?
(a) 30 × 2⁽ⁿ⁾ - 1
(b) 30 × (2^{(n - 1)} - 1)
(c) 30 × (2^{(n + 1)} - 1)
(d) 30 × 2ⁿ - 30

Answer

The number of bacteria in a certain culture doubles every hour. this forms G.P.

30, 60, 120, 240...

1. a = 30

r = $\dfrac{60}{30}$ = 2

n = 5 [At the end of 4th hour, means starting of 5th hour]

We know that,

T_n = ar^{n - 1}

Substituting values we get,

T₅ = 30(2)^{5 - 1}

= 30.(2)⁴

= 480.

Hence, option (b) is the correct option.

2. Given,

n = 11 [At the end of 10th hour, means starting of 11th hour]

a = 30

r = 2

We know that,

T_n = ar^{n - 1}

⇒ T₁₁ = 30(2)^{11 - 1}

= 30.(2)¹⁰

= 30.(1024)

= 30720.

Hence, option (a) is the correct option.

3. Given,

a = 30

r = 2

n = n + 1 [At the end of nth hour, means starting of (n + 1) th hour]

We know that,

⇒ T_n = ar^{n - 1}

⇒ T_n = 30.(2)^{n + 1 - 1}

= 30 × (2)ⁿ.

Hence, option (c) is the correct option.

4. Formula for sum of n terms of a G.P.,

$S_n = \dfrac{a(r^n-1)}{r-1}$ [r > 1]

Given,

a = 30

r = 2

n = 5

$\Rightarrow S_5 = \dfrac{30(2^5 - 1)}{2-1} \\[1em] = \dfrac{30(32 - 1)}{1} \\[1em] = 30.(31) \\[1em] = 930.$

Hence, option (b) is the correct option.

5. Formula for sum of n terms of a G.P.

$S_n = \dfrac{a(r^n-1)}{r-1}$ [r > 1]

Given,

a = 30

r = 2

$\Rightarrow S_5 = \dfrac{30(2^n - 1)}{2-1} \\[1em] = \dfrac{30(2^n - 1)}{1} \\[1em] = 30.(2^n - 1) \\[1em] = 30 \times 2^n - 30 .$

Hence, option (d) is the correct option.

Question 2

Case Study

A man writes a letter to four of his friends. He asks each one of them to copy the letter and mail it to four different persons with the instruction that they move the chain similarly. Assume that the chain is not broken and it costs ₹4 to mail one letter.

Based on the above given information, answer the following questions:

1.The number of letters mailed in the 6th set of letters is :
(a) 2048
(b) 8192
(c) 4096
(d) 1024

2.The amount spent on the postage of 6th set of letters is :
(a) ₹ 16,384
(b) ₹ 8,192
(c) ₹ 32,768
(d) ₹ 4,096

3.The total number of letters mailed till the 5th set of letters is :
(a) 1024
(b) 4092
(c) 1236
(d) 1364

4.The amount spent on the postage till the 5th set of letters is :
(a) ₹ 16,368
(b) ₹ 4,096
(c) ₹ 5,456
(d) ₹ 4,944

5.The amount spent on the postage till the nth set of letters is :
(a) ₹ $\Big(\dfrac{4}{3}(4^n - 4)\Big)$

(b) ₹ $\Big(\dfrac{16}{3}(4^n - 1)\Big)$

(c) ₹ $\Big(\dfrac{2}{3}(4^n - 1)\Big)$

(d) ₹ $\Big(16(4^n - 1)\Big)$

Answer

Since the number of letters mailed increases by a factor of 4 at every step, the sequence of letters mailed forms a Geometric Progression.

4,16,64.....

1. In the given G.P.,

a = 4

r = 4

We know that,

T_n = ar^{n - 1}

⇒ T₆ = 4.(4)^{6 - 1}

= 4(4)⁵

= 4(1024)

= 4096.

Hence, option (c) is the correct option.

2. In the 6th set, 4096 letters are mailed and each letter costs ₹ 4.

Amount spent on the postage of 6th set of letters = 4096 × ₹ 4 = ₹ 16,384.

Hence, option (a) is the correct option.

3. Formula for sum of n terms of a G.P.

$S_n = \dfrac{a(r^n-1)}{r-1}$ [r > 1]

Given,

a = 4

r = 4

n = 5

$\Rightarrow S_5 = \dfrac{4(4^5 - 1)}{4-1} \\[1em] = \dfrac{4(4^5 - 1)}{3} \\[1em] = \dfrac{4(1024 - 1)}{3} \\[1em] = \dfrac{4(1023)}{3} \\[1em] = \dfrac{4092}{3} \\[1em] = 1364.$

Hence, option (d) is the correct option.

4. Formula for sum of n terms of a G.P.

$S_n = \dfrac{a(r^n-1)}{r-1}$ [r > 1]

Given,

a = 4

r = 4

n = 5

$\Rightarrow S_5 = \dfrac{4(4^5 - 1)}{4-1} \\[1em] = \dfrac{4(4^5 - 1)}{3} \\[1em] = \dfrac{4(1024 - 1)}{3} \\[1em] = \dfrac{4(1023)}{3} \\[1em] = \dfrac{4092}{3} \\[1em] = 1364.$

Amount spent till the 5th set of letters postage = 1364 × ₹ 4 = ₹ 5,456

Hence, option (c) is the correct option.

5. Formula for sum of n terms of a G.P.

$S_n = \dfrac{a(r^n-1)}{r-1}$ [r > 1]

Given,

a = 4

r = 4

$\Rightarrow S_5 = \dfrac{4(4^n - 1)}{4-1} \\[1em] = \dfrac{4(4^n - 1)}{3} \\[1em]$

Amount spent till the nth set of letters postage = $\dfrac{4(4^n - 1)}{3}$ × ₹ 4 = $₹ \dfrac{16}{3}(4^n - 1)$ .

Hence, option (b) is the correct option.

Assertion Reason Questions

Question 1

Assertion (A): The nth term of a G.P. is given by T_n = ar^{n − 1}.

Reason (R): A sequence a₁, a₂, a₃, ....... is said to be a G.P. if $\dfrac{a_2}{a_1} = \dfrac{a_3}{a_2} = \dfrac{a_4}{a_3}$ = constant known as the common ratio.

A is true, R is false.
A is false, R is true.
Both A and R are true.
Both A and R are false.

Answer

The nth term of a G.P. is given by :

T_n = ar^{n - 1}

Assertion (A) is true.

A sequence $a_1, a_2, a_3, \dots$ is said to be a G.P. if $\dfrac{a_1}{a_2} = \dfrac{a_2}{a_3} = \dfrac{a_3}{a_4} = \text{constant}$ known as the common ratio.

A sequence is a G.P. if the ratio between any consecutive terms is a constant value. This constant is called the common ratio.

Reason is true.

Both A and R are true.

Hence, option 3 is the correct option.

Question 2

Assertion (A): The 50th term from the end of the G.P. 4, 6, 9, $\dfrac{27}{2}, ......\dfrac{6561}{64}$ is $\dfrac{27}{2}$ .

Reason (R): nth term from the end of a G.P. is given by $\dfrac{l}{r^{n-1}}$ .

A is true, R is false
A is false, R is true
Both A and R are true
Both A and R are false

Answer

Given,

a = 4

n = 50

r = $\dfrac{6}{4} = \dfrac{3}{2}$

l = $\dfrac{6561}{64}$

We know that,

$\text{nth term from end} = \dfrac{l}{r^{n - 1}}$

Substituting values we get,

$\Rightarrow \text{50th term from end} = \dfrac{\dfrac{6561}{64}}{\Big(\dfrac{3}{2}\Big)^{50-1}} \\[1em] = \dfrac{\dfrac{3^8}{2^6}}{\dfrac{3^{49}}{2^{49}}} \\[1em] = \dfrac{3^8}{2^6} \times \dfrac{2^{49}}{3^{49}} \\[1em] = 2^{49 - 6} \times 3 ^{8 - 49} \\[1em] = 2^{43} \times 3^{-41}$

= $\dfrac{2^{43}}{3^{41}}$ ≠ $\dfrac{27}{2}$ .

So, Assertion is false.

nth term from the end of a G.P. is given by: $\dfrac{l}{r^{n-1}}$

So, reason is true.

Hence, option 2 is the correct option.

Question 3

Assertion (A): The sum of first 10 terms of the G.P. 3, 6, 9, 12, ...... is 3096.

Reason (R): The sum of first n-terms of a G.P. is given by $S_n = \dfrac{a(r^n + 1)}{r + 1}$ .

A is true, R is false
A is false, R is true
Both A and R are true
Both A and R are false

Answer

Given,

3, 6, 9, 12, …

Since, $\dfrac{6}{3}$ ≠ $\dfrac{9}{6}$

Thus, r is not constant. Hence it is not G.P.

Assertion (A) is false.

The correct formula for the sum of n terms of a G.P.

$S_n = \dfrac{a(r^n - 1)}{r - 1}$ [r > 1]

Reason (R) is false

Both A and R are false.

Hence, option 4 is the correct option.

Chapter 11

Geometric Progression

Class - 10 RS Aggarwal Mathematics Solutions

Exercise 11A

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16

Exercise 11B

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Multiple Choice Questions

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18

Question 19

Question 20

Question 21

Question 22

Question 23

Question 24

Question 25

Question 26

Question 27

Question 28

Question 29

Question 30

Question 31

Question 32

Question 33

Question 34

Case Study Based Questions

Question 1

Question 2

Assertion Reason Questions

Question 1

Question 2

Question 3