Geometric Progression

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

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

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

a = 2

r = $\frac{6}{2}$ = 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.

Given,

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

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

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

a = 625

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

We know that,

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

T_n = ar^{n - 1}

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

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

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

= 5^{5 - n}

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

10th term,

T₁₀ = 5^{5 - 10}

= 5^-5

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

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

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

Given,

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

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

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

a = -27

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

We know that,

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

T_n = ar^{n - 1}

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

9th term

$$\begin{aligned}\Rightarrow T_9 = \frac{1}{-3^{9 - 4}} \\[1em] = \frac{1}{-3^{5}} \\[1em] = -\frac{1}{243} \\[1em]\end{aligned}$$

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

Given,

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

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

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

a = 2

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

We know that,

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

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

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

Given,

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

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

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

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

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

We know that,

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

T_n = ar^{n - 1}

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

6th term,

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

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

Given,

0.4, 0.8, 1.6,.....

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

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

a = 0.4

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

We know that,

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

T_n = ar^{n - 1}

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

7th term,

T₇ = $\frac{2^{7}}{5}$

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

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

Given,

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

a = 3

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

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}

⇒ $\frac{768}{3}$ = (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.

Given,

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

a = 5

r = $\frac{10}{5}$ = 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}

⇒ $\frac{640}{5}$ = 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.

Given,

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

a = $\sqrt{3}$

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

T_n = 729

We know that,

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

T_n = ar^{n - 1}

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

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

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) :

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

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

⇒ a(2)⁴ = 80

⇒ 16a = 80

⇒ a = $\frac{80}{16}$

⇒ 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, ......

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) :

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

Substituting r = 2 in Equation 1, we get :

⇒ a(2)³ = 16

⇒ a(8) = 16

⇒ a = $\frac{16}{8}$

⇒ a = 2.

Hence, a = 2 and r = 2.

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

Given,

First term (a) = 24

Fifth term (ar⁴) = 8

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

By formula,

Ninth term of G.P. (a₉) = ar⁸

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

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

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 $\frac{1}{18}$

⇒ ar^{4 - 1} = $\frac{1}{18}$

⇒ ar³ = $\frac{1}{18}$ ....(1)

Given,

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

⇒ ar^{7 - 1} = $-\frac{1}{486}$

⇒ ar⁶ = $-\frac{1}{486}$....(2)

Divide Equation 2 by Equation 1:

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

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

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

G.P. is,

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

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

We know that,

In G.P. we have constant common ratio.

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

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.

Given,

a = 16

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

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

We know that,

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

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

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

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

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

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

l = 54

We know that,

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

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

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

Hence, 4th term from end = 2.

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:

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

⇒ 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 = $\frac{10}{8}$

⇒ a = $\frac{5}{4}$.

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

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

⇒ 2^{n - 1} = $\frac{640 × 4}{5}$

⇒ 2^{n - 1} = 128 × 4

⇒ 2^{n - 1} = 512

⇒ 2^{n - 1} = 2⁹

⇒ n - 1 = 9

⇒ n = 9 + 1

⇒ n = 10.

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

Given,

a = 1

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

n = 8

We know that,

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

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

Substituting values we get :

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

Hence, S₈ = 3280.

Given,

a = 1

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

n = 10

We know that,

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

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

Substituting values we get :

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

Rationalizing the Denominator :

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

Hence, S₁₀ = $121(\sqrt3 + 1)$.

Given,

a = 1

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

n = 9

We know that,

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

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

Substituting values we get :

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

Hence, S₉ = $\frac{171}{256}$.

Given,

a = 0.1

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

n = 6

We know that,

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

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

Substituting values we get :

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

Hence, S₆ = 0.111111.

Given,

a = 1

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

n = 6

We know that,

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

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

Substituting values we get :

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

Hence, S₆ = $\frac{182}{243}$.

Given,

a = 27

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

We know that,

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

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

Substituting values we get :

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

Hence, S₇ = $\frac{1093}{27}$.

Given,

⇒ T₄ = $\frac{1}{27}$

⇒ ar³ = $\frac{1}{27}$ .....(1)

Given,

⇒ T₇ = $\frac{1}{729}$

⇒ ar⁶ = $\frac{1}{729}$ ..........(2)

Dividing Equation (2) by Equation (1) :

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

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

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

We know that,

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

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

Substituting values we get :

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

Hence, S₆ = $\frac{364}{243}$.

(i) Let the first term of the Geometric Progression be a and the common ratio be r.

By formula,

T_n = arⁿ

Given,

The 6th term is 32.

⇒ ar^{6 - 1} = 32

⇒ ar⁵ = 32 .........(1)

The 9th term is 256.

⇒ ar^{9 - 1} = 256

⇒ ar⁸ = 256 .........(2)

Divide equation (2) by (1), we get:

$$\begin{aligned}\Rightarrow \frac{a \times r^8}{a \times r^5} = \frac{256}{32} \\[1em] \Rightarrow r^{8 - 5} = 8 \\[1em] \Rightarrow r^{3} = 8 \\[1em] \Rightarrow r = \sqrt[3]{8} \\[1em] \Rightarrow r = 2.\end{aligned}$$

Substitute value of r in equation (1), we get:

$$\begin{aligned}\Rightarrow a \times 2^5 = 32 \\[1em] \Rightarrow a \times 32 = 32 \\[1em] \Rightarrow a = \frac{32}{32} \\[1em] \Rightarrow a = 1.\end{aligned}$$

Hence, the first term = 1 and common ratio = 2.

(ii) By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow S_{10} = \frac{1(2^{10} - 1)}{2 - 1} \\[1em] \Rightarrow S_{10} = \frac{1024 - 1}{1} \\[1em] \Rightarrow S_{10} = 1023.\end{aligned}$$

Hence, sum of the first 10 terms of the G.P. = 1023.

Given,

G.P. : 15, 30, 60, 120 ......

First term (a) = 15

Common ratio (r) = $\frac{30}{15}$ = 2

(i) nth term of G.P. = ar^{n - 1} = 15 x 2^{n - 1}

= $\frac{15}{2}$ x 2ⁿ

= 7.5 x 2ⁿ

Hence, nth term of the given G.P. is 7.5 x 2ⁿ

(ii) Let sum of n terms of G.P. is 945.

By formula,

Sum of n terms of G.P. = $\frac{a(r^n - 1)}{(r - 1)}$

Substituting values we get :

$$\begin{aligned}\Rightarrow 945 = \frac{15.(2^n - 1)}{2 - 1} \\[1em] \Rightarrow 945 = \frac{15.(2^n - 1)}{1} \\[1em] \Rightarrow 2^n - 1 = \frac{945}{15} \\[1em] \Rightarrow 2^n - 1 = 63 \\[1em] \Rightarrow 2^n = 63 + 1 \\[1em] \Rightarrow 2^n = 64 \\[1em] \Rightarrow 2^n = 2^6 \\[1em] \Rightarrow n = 6.\end{aligned}$$

Hence, sum of 6 terms of G.P. = 945.

In the given G.P.,

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

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

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

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

We know that,

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

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

Substituting values we get :

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

Hence, n = 5.

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

Given,

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

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

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 = $\frac{80}{100}$ [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,

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

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

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

Now,

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

a = 10

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

We know that,

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

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

$$\begin{aligned}\Rightarrow S_n = \frac{10(10^n - 1)}{10 - 1} \\[1em] = \frac{10(10^n - 1)}{9}.\end{aligned}$$

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

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

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

Given,

mth term = 2^m + 2m

1st term = 2¹ + 2 × 1

2nd term = 2² + 2 × 2

nth term = 2ⁿ + 2 × n

$$\begin{aligned}S_n = [(2^1 + 2 \times 1) + (2^2 + 2 \times 2) +(2^3 + 2 \times 3)+........ + (2^n + 2 \times n)] \\[1em] = [2^1 + 2^2 + 2^3 + ..... 2^n + (2 \times 1 + 2 \times 2 + 2 \times 3 + ..... + 2 \times n)] \\[1em] = (2^1 + 2^2 + 2^3 + ..... + 2^n) + 2(1 + 2 + 3 + ..... + n)......(1)\end{aligned}$$

Calculating :

2¹ + 2² + ........ + 2ⁿ

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

By using formula,

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

Substitute values, we get:

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

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

Substitute values, we get:

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

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

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

Substitute values in (1) we get,

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

= 2(2ⁿ - 1) + n(n + 1)

Hence, S_n = 2(2ⁿ - 1) + n(n + 1).

In the series,

2, 0, 4, 0, 8, 0

$\frac{0}{2} \ne \frac{4}{0}$

∴ 2, 0, 4, 0, 8, 0 is not a G.P.

Hence, Option 4 is the correct option.

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. $\frac{1}{r}$.

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

Hence, option 2 is the correct option.

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.

We know that,

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

T_n = ar^{n - 1}

In the given G.P.,

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

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

⇒ T_n = $\frac{1}{4}$ (-2)^{n - 1}

= $\frac{1}{2^2}$.(-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.

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 = $\frac{4}{2}$ = 2

n = 12

⇒ T₁₂ = 2.(2)^{12 - 1}

= 2(2)¹¹

= (2)^{11 + 1}

= (2)¹²

= 4096.

Hence, option 3 is the correct option.

In the given A.P.,

a = $\sqrt3$

r = $\frac{3\sqrt3}{\sqrt3}$ = 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.

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.

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

Hence, option 4 is the correct option.

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

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

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

Hence, option 2 is the correct option.

$$\begin{aligned}r = \frac{ax}{\frac{1}{a^3x^3}} \\[1em] = ax \times a^3x^3 = a^4x^4.\end{aligned}$$

Hence, option 4 is the correct option.

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

= 0.1

Hence, option 1 is the correct option.

In the given G.P.,

a = x³

r = $\frac{x^5}{x^3}$ = 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.

In the given G.P.,

a = 1

r = $\frac{-a}{1}$ = -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.

In the given G.P.,

a = 3

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

Hence, option 2 is the correct option.

In the given G.P.,

a = 2

r = $\frac{8}{2}$ = 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} = $\frac{131072}{2}$

⇒ 2^{2n - 2} = 65536

⇒ 2^{2n - 2} = 2¹⁶

Equate exponents:

⇒ 2n - 2 = 16

⇒ 2n = 16 + 2

⇒ 2n = 18

⇒ n = $\frac{18}{2}$ = 9.

Hence, option 1 is the correct option.

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.

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.

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.

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

⇒ (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 = $\frac{-3a}{-a}$ = 3.

Hence, option 2 is the correct option.

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

a = 8

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

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

We know that,

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

Substitute values we get:

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

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

Hence, option 1 is the correct option.

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

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

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

l = 162.

We know that,

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

Substitute values we get:

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

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

Hence, option 3 is the correct option.

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

The product of the three terms is given as -1

⇒ $\frac{a}{r}$ × a × ar = -1

⇒ a³ = -1

⇒ a = $\sqrt[3]{-1}$

⇒ a = -1.

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

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

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

The sum of the three terms is :

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

Hence, option 4 is the correct option.

We know that,

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

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

Hence, option 3 is the correct option.

In the given G.P.,

a = 1

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

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

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

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

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

Hence, option 3 is the correct option.

In the given G.P.,

a = 3

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

n = 7

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

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

Substituting values we get :

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

Hence, option 3 is the correct option.

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 = $-\frac{4}{3}$.

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.

Given,

x, y and z are in G.P.

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

$$\begin{aligned}\Rightarrow \frac{y}{x} = \frac{z}{y} \\[1em] \Rightarrow y^2 = xz \\[1em] \Rightarrow y = \sqrt{xz}.\end{aligned}$$

Hence, option 4 is the correct option.

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.

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.

Given,

G.P. a, ar, ar², ........, l.

We know that,

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

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

Hence, option 1 is the correct option.

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 $\frac{a_2}{a_1} = \frac{a_3}{a_2} = \frac{a_4}{a_3} = \text{constant}$ known as the common ratio.

Reason (R) is false.

Hence, option 3 is the correct option.

Given,

a = 4

n = 50

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

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

We know that,

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

Substituting values we get,

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

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

So, Assertion is false.

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

So, Reason (R) is true.

Hence, option 4 is the correct option.

Given,

3, 6, 9, 12, …

Since, $\frac{6}{3}$ ≠ $\frac{9}{6}$

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.

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

Reason (R) is true.

Hence, option 4 is the correct option.

Terms of the sequence given by $(\sqrt3)^n$ are :

$(\sqrt3)^1, (\sqrt3)^2, (\sqrt3)^3, (\sqrt3)^4, (\sqrt3)^5, .......$

$\sqrt3, 9, 3\sqrt3, 9, 9\sqrt3, .......$

(a) Ratio between terms = $\frac{3}{\sqrt3} = \sqrt3$

Hence, the sequence is a G.P. with common ratio = $\sqrt3$.

(b) By formula,

Sum of G.P. (S) = $\frac{a(r^n - 1)}{(r - 1)}$

Substituting values we get :

$$\begin{aligned}\Rightarrow S_{10} = \frac{\sqrt3[(\sqrt3)^10 - 1)}{(\sqrt3 - 1)} \\[1em] \Rightarrow p(3 + \sqrt3) = \frac{\sqrt3[(\sqrt3)^10 - 1]}{(\sqrt3 - 1)} \\[1em] \Rightarrow p = \frac{\sqrt3(243 - 1)}{(\sqrt3 - 1)(3 + \sqrt3)} \\[1em] \Rightarrow p = \frac{\sqrt3(242)}{(\sqrt3 - 1)(3 + \sqrt3)} \\[1em] \Rightarrow p = \frac{\sqrt3(242)}{(\sqrt3 - 1)\sqrt3(\sqrt3 + 1)} \\[1em] \Rightarrow p = \frac{242}{(\sqrt3)^2 - (1)^2} \\[1em] \Rightarrow p = \frac{242}{3 - 1} \\[1em] \Rightarrow p = \frac{242}{2} = 121\end{aligned}$$

Hence, p = 121.

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

30, 60, 120, 240...

1. a = 30

r = $\frac{60}{30}$ = 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.,

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

Given,

a = 30

r = 2

n = 5

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

Hence, option (b) is the correct option.

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

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

Given,

a = 30

r = 2

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

Hence, option (d) is the correct option.

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.

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

Given,

a = 4

r = 4

n = 5

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

Hence, option (d) is the correct option.

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

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

Given,

a = 4

r = 4

n = 5

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

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.

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

Given,

a = 4

r = 4

$$\begin{aligned}\Rightarrow S_5 = \frac{4(4^n - 1)}{4-1} \\[1em] = \frac{4(4^n - 1)}{3} \\[1em]\end{aligned}$$

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

Hence, option (b) is the correct option.