Geometric Progression — Exercise 11(A)

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.

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

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 = $-\dfrac{1}{243}$.

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.

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 = $\dfrac{8}{81}$.

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

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.

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.

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.

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, ......

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.

In a Geometric Progression (G.P.) the first term is 24 and the fifth term is 8. Find the ninth term of the G.P.

Answer

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

Given,

First term (a) = 24

Fifth term (ar⁴) = 8

$\Rightarrow \dfrac{ar^4}{a} = \dfrac{8}{24} \\[1em] \Rightarrow r^4 = \dfrac{1}{3} \Rightarrow (r^4)^2 = \Big(\dfrac{1}{3}\Big)^2 \\[1em] \Rightarrow r^8 = \dfrac{1}{9}.$

By formula,

Ninth term of G.P. (a₉) = ar⁸

= $24 \times \dfrac{1}{9} = \dfrac{8}{3}$.

Hence, ninth term of G.P. = $\dfrac{8}{3}$.

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 $-\dfrac{3}{2}, \dfrac{1}{2}, -\dfrac{1}{6}, ........$

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.

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 = $\dfrac{1}{16}$.

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.

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.