Geometric Progression

Given,

First term of G.P. = 8

Common ratio = -2

By formula,

⇒ a_n = ar^{n - 1}

Substituting values we get :

⇒ a₃ = 8 × (-2)^{3 - 1}

⇒ a₃ = 8 × (-2)²

⇒ a₃ = 8 × 4

⇒ a₃ = 32.

Hence, Option 3 is the correct option.

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

By formula,

⇒ a_n = ar^{n - 1}

Given,

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

⇒ a₄ = 16

⇒ ar^{4 - 1} = 16

⇒ ar³ = 16 .........(1)

Given,

7^th term of a G.P. is 128.

⇒ a₇ = 128

⇒ ar^{7 - 1} = 128

⇒ ar⁶ = 128 .........(2)

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

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

Hence, Option 1 is the correct option.

By formula,

Common ratio = $\frac{a_{n + 1}}{a_n}$

Given,

(2x + 2) and (3x + 3) are two consecutive terms of a G.P.

$$\begin{aligned}\therefore \text{Common ratio } = \frac{3x + 3}{2x + 2} \\[1em] = \frac{3(x + 1)}{2(x + 1)} \\[1em] = \frac{3}{2}.\end{aligned}$$

Hence, Option 3 is the correct option.

Let first five terms of G.P. be

$\frac{a}{r^2}, \frac{a}{r}, a, ar, ar^2$

Given,

3rd term of G.P. is 3.

∴ a = 3

Product of first five terms are

$$\begin{aligned}\Rightarrow \frac{a}{r^2} \times \frac{a}{r} \times a \times ar \times ar^2 \\[1em] \Rightarrow a^5 \\[1em] \Rightarrow 3^5.\end{aligned}$$

Hence, Option 3 is the correct option.

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

By formula,

⇒ a_n = ar^{n - 1}

Given,

8^th term of a G.P. is 192.

⇒ a₈ = 192

⇒ ar^{8 - 1} = 192

⇒ ar⁷ = 192

Substituting value of common ratio (r) = 2 in above equation, we get :

⇒ a(2)⁷ = 192

⇒ 128a = 192

⇒ a = $\frac{192}{128} = \frac{3}{2}$.

Hence, Option 4 is the correct option.

Since,

$\frac{4}{1} = \frac{16}{4} = 4$.

Hence, the above sequence is a G.P. with r = 4 and a = 1.

We know that n^th term of G.P.,

a_n = ar^{n - 1}

a₉ = 1.(4)^{9 - 1}

= (4)⁸

= 65536.

Hence, 9^th term of the series = 65536.

Since,

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

Hence, the above sequence is a G.P. with r = $\sqrt{3}$ and a = 1.

We know that n^th term of G.P.,

a_n = ar^{n - 1}

a₇ = 1.$(\sqrt{3})^{7 - 1}$

= $(\sqrt{3})^6$

= 27.

Hence, 7^th term of the G.P. = 27.

Sequence = $\frac{3}{4}, \frac{3}{2}, 3, ........$

Calculating ratio between terms,

$\frac{\frac{3}{2}}{\frac{3}{4}} = \frac{3 \times 4}{2 \times 3} = 2, \frac{3}{\frac{3}{2}} = 2$.

Since,

$\frac{\frac{3}{2}}{\frac{3}{4}} = \frac{3}{\frac{3}{2}} = 2$.

Hence, the sequence is a G.P. with r = 2 and a = $\frac{3}{4}$

We know that n^th term of G.P.,

a_n = ar^{n - 1}

a₈ = $\frac{3}{4} \times (2)^{8 - 1}$

= $\frac{3}{4} \times 2^7$

= $\frac{3}{4} \times 128$

= 3 × 32

= 96.

Hence, a₈ = 96.

Since,

$\frac{5}{\sqrt{5}} = \frac{5\sqrt{5}}{5} = \sqrt{5}$

Hence, the sequence $\sqrt{5}, 5, 5\sqrt{5}, .......$ is a G.P. with r = $\sqrt{5} \text{ and } a = \sqrt{5}.$

Next three terms are = 4^th, 5^th and 6^th.

We know that n^th term of G.P.,

a_n = ar^{n - 1}

⇒ a₄ = ar^{(4 - 1)}

= ar³

= $\sqrt{5}(\sqrt{5})^3 = \sqrt{5}(5\sqrt{5})$

= 25.

⇒ a₅ = ar^{(5 - 1)}

= ar⁴

= $\sqrt{5}(\sqrt{5})^4 = \sqrt{5}(25)$

= $25\sqrt{5}$.

⇒ a₆ = ar^{(6 - 1)}

= ar⁵

= $\sqrt{5}(\sqrt{5})^5 = \sqrt{5}(25\sqrt{5})$

= 125.

Hence, next three terms of the G.P. are = 25, 25√5 and 125.

Rationalising the term, $\frac{\sqrt{3} - 1}{2}$ we get,

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

So, Sequence = $\sqrt{3} + 1, 1, \frac{1}{\sqrt{3} + 1},$ ...........

Common ratio(r) = $\frac{1}{\sqrt{3} + 1}$.

We know that n^th term of G.P.,

a_n = ar^{n - 1}

$$\begin{aligned}\Rightarrow a_7 = (\sqrt{3} + 1)\Big(\frac{1}{\sqrt{3} + 1}\Big)^{7 - 1} \\[1em] = (\sqrt{3} + 1)\Big(\frac{1}{\sqrt{3} + 1}\Big)^{6} \\[1em] = \Big(\frac{1}{\sqrt{3} + 1}\Big)^{5} \\[1em] = \Big(\frac{1}{\sqrt{3} + 1} \times \frac{\sqrt{3} - 1}{\sqrt{3} - 1}\Big)^{5} \\[1em] = \Big(\frac{\sqrt{3} - 1}{\sqrt{3}^2 - (1)^2}\Big)^5 \\[1em] = \Big(\frac{\sqrt{3} - 1}{3 - 1}\Big)^5 \\[1em] = \Big(\frac{\sqrt{3} - 1}{2}\Big)^5 \\[1em] = \frac{1}{32}(\sqrt{3} - 1)^5.\end{aligned}$$

Hence, seventh term of the G.P. = $\frac{1}{32}(\sqrt{3} - 1)^5.$

Since,

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

Hence, the above series is a G.P. with r = -3 and a = 2.

We know that n^th term of G.P.,

a_n = ar^{n - 1}

Next two terms of the series are 5^th and 6^th.

⇒ a₅ = ar⁴

= 2.(-3)⁴

= 2 × 81

= 162.

⇒ a₆ = ar⁵

= 2.(-3)⁵

= 2 × -243

= -486.

Hence, the next two terms are 162 and -486.

Common ratio (r) = $\frac{\frac{5}{\sqrt{3}}}{-10} = -\frac{5}{10\sqrt{3}} = -\frac{1}{2\sqrt{3}}$.

Let n^th term of G.P. be $-\frac{5}{72}$.

$$\begin{aligned}\therefore ar^{n - 1} = -\frac{5}{72} \\[1em] \Rightarrow -10 \times \Big(-\frac{1}{2\sqrt{3}}\Big)^{n - 1} = -\frac{5}{72} \\[1em] \Rightarrow \Big(-\frac{1}{2\sqrt{3}}\Big)^{n - 1} = -\frac{5}{72} \times -\frac{1}{10} \\[1em] \Rightarrow \Big(-\frac{1}{2\sqrt{3}}\Big)^{n - 1} = \frac{1}{144} \\[1em] \Rightarrow \Big(-\frac{1}{2\sqrt{3}}\Big)^{n - 1} = \Big(-\frac{1}{2\sqrt{3}}\Big)^4 \\[1em] \Rightarrow n - 1 = 4 \\[1em] \Rightarrow n = 5.\end{aligned}$$

Hence, 5^th term of the G.P. is $-\frac{5}{72}.$

Let first term of the G.P. be a and it's common ratio be r.

Given,

⇒ a₅ = 81

⇒ ar⁴ = 81 ........(i)

Also,

⇒ a₂ = 24

⇒ ar = 24 ........(ii)

Dividing (i) by (ii) we get,

$$\begin{aligned}\Rightarrow \frac{ar^4}{ar} = \frac{81}{24} \\[1em] \Rightarrow r^3 = \frac{27}{8} \\[1em] \Rightarrow r^3 = \Big(\frac{3}{2}\Big)^3 \\[1em] \Rightarrow r = \frac{3}{2}.\end{aligned}$$

Substituting value of r in (ii) we get,

$$\begin{aligned}\Rightarrow a \times \frac{3}{2} = 24 \\[1em] \Rightarrow a = \frac{2}{3} \times 24 \\[1em] \Rightarrow a = 16.\end{aligned}$$

⇒ a₃ = ar²

= $16 \times \Big(\frac{3}{2}\Big)^2$

= $16 \times \frac{9}{4}$

= 4 × 9

= 36.

⇒ a₄ = ar³

= $16 \times \Big(\frac{3}{2}\Big)^3$

= $16 \times \frac{27}{8}$

= 2 × 27

= 54.

G.P. = 16, 24, 36, 54, 81, ...........

Hence, G.P. = 16, 24, 36, 54, 81, ...........

Let first term of the G.P. be a and it's common ratio be r.

Given,

⇒ a₄ = $\frac{1}{18}$

⇒ ar³ = $\frac{1}{18}$ ........(i)

Also,

⇒ a₇ = $-\frac{1}{486}$

⇒ ar⁶ = $-\frac{1}{486}$ ........(ii)

Dividing (ii) by (i) we get,

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

Substituting value of r in (i) we get,

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

⇒ a₂ = ar

= $-\frac{3}{2} \times \Big(-\frac{1}{3}\Big)$

= $\frac{1}{2}$.

⇒ a₃ = ar²

= $-\frac{3}{2} \times \Big(-\frac{1}{3}\Big)^2$

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

= $-\frac{1}{6}.$

G.P. = $-\frac{3}{2}, \frac{1}{2}, -\frac{1}{6}, \frac{1}{18} ...........$

Hence, G.P. = $-\frac{3}{2}, \frac{1}{2}, -\frac{1}{6}, \frac{1}{18} ...........$

Let first term of the G.P. be a and it's common ratio be r.

Given,

⇒ a = 2

⇒ a₃ = 8

⇒ ar² = 8

⇒ 2r² = 8

⇒ r² = 4

⇒ r =√4 = ±2

a₂ = ar

Let r = -2

⇒ a₂ = 2(-2) = -4

Let r = 2

⇒ a₂ = 2(2) = 4.

Hence, a₂ = 4 or -4.

Let first term of the G.P. be a and it's common ratio be r.

Given,

⇒ a₃.a₈ = 243

⇒ ar².ar⁷ = 243

⇒ a²r⁹ = 243 ..........(i)

Also,

⇒ a₄ = 3

⇒ ar³ = 3

⇒ a = $\frac{3}{r^3}$ ........(ii)

Substituting value of a from (ii) in (i) we get,

$$\begin{aligned}\Rightarrow \Big(\frac{3}{r^3}\Big)^2 \times r^9 = 243 \\[1em] \Rightarrow \frac{9}{r^6} \times r^9 = 243 \\[1em] \Rightarrow 9r^3 = 243 \\[1em] \Rightarrow r^3 = 27 \\[1em] \Rightarrow r = \sqrt[3]{27} \\[1em] \Rightarrow r = 3.\end{aligned}$$

Substituting value of r in (ii),

$$\begin{aligned}\Rightarrow a = \frac{3}{r^3} \\[1em] = \frac{3}{3^3} \\[1em] = \frac{1}{3^2} \\[1em] = \frac{1}{9}.\end{aligned}$$

a₇ = ar⁶

= $\frac{1}{9} \times 3^6$

= $\frac{1}{9} \times 729$

= 81.

Hence, 7^th term = 81.

Let first term of the G.P. be a and it's common ratio be r.

Given,

⇒ a₄ = 54

⇒ ar³ = 54 ........(i)

Also,

⇒ a₇ = 1458

⇒ ar⁶ = 1458 ........(ii)

Dividing (ii) by (i) we get

$$\begin{aligned}\Rightarrow \frac{ar^6}{ar^3} = \frac{1458}{54} \\[1em] \Rightarrow r^3 = 27 \\[1em] \Rightarrow r = \sqrt[3]{27} \\[1em] \Rightarrow r = 3.\end{aligned}$$

Substituting value of r in (i) we get,

⇒ a(3)³ = 54

⇒ 27a = 54

⇒ a = $\frac{54}{27}$ = 2.

a₂ = ar

= 2.(3) = 6.

a₃ = ar²

= 2.(3)²

= 2.(9) = 18.

G.P. = 2, 6, 18, 54, ......

Hence, G.P. = 2, 6, 18, 54, ......

Let first term of the G.P. be a and it's common ratio be r.

Given,

⇒ a₂ = 6

⇒ ar = 6 ........(i)

Also,

⇒ a₅ = 9a₃

⇒ ar⁴ = 9ar²

⇒ $\frac{\text{ar}^4}{\text{ar}^2}$ = 9

⇒ r² = 9

⇒ r = √9

⇒ r = ±3

As all terms of G.P. are positive so, r ≠ -3

∴ r = 3

Substituting r in (i),

⇒ 3a = 6

⇒ a = 2.

G.P. = a, ar, ar², ar³, ......

= 2, 6, 18, 54, .......

Hence, G.P. = 2, 6, 18, 54, .......

Let first term of the G.P. be a and it's common ratio be r.

Given,

⇒ a₄ = 10

⇒ ar³ = 10 ........(i)

Also,

⇒ a₇ = 80

⇒ ar⁶ = 80 .........(ii)

Dividing (ii) by (i) we get,

$$\begin{aligned}\Rightarrow \frac{ar^6}{ar^3} = \frac{80}{10} \\[1em] \Rightarrow r^3 = 8 \\[1em] \Rightarrow r = \sqrt[3]{8} \\[1em] \Rightarrow r = 2.\end{aligned}$$

Substituting r in (i) we get,

$$\begin{aligned}\Rightarrow a(2)^3 = 10 \\[1em] \Rightarrow 8a = 10 \\[1em] \Rightarrow a = \frac{10}{8} \\[1em] \Rightarrow a = \frac{5}{4}.\end{aligned}$$

Let n be no. of terms,

ar^{n - 1} = 2560

$$\begin{aligned}\Rightarrow \frac{5}{4}\times (2)^{n - 1} = 2560 \\[1em] \Rightarrow (2)^{n - 1} = \frac{4}{5} \times 2560 \\[1em] \Rightarrow (2)^{n - 1} = 4 \times 512 \\[1em] \Rightarrow (2)^{n - 1} = 2048 \\[1em] \Rightarrow (2)^{n - 1} = (2)^{11} \\[1em] \Rightarrow n - 1 = 11 \\[1em] \Rightarrow n = 12.\end{aligned}$$

Hence, first term = $\frac{5}{4}$, common ratio = 2 and number of terms = 12.

Let first term of the G.P. be a and it's common ratio be r.

Given,

⇒ a₄ = 54

⇒ ar³ = 54 ........(i)

Also,

⇒ a₉ = 13122

⇒ ar⁸ = 13122 .........(ii)

Dividing (ii) by (i) we get,

$$\begin{aligned}\Rightarrow \frac{ar^8}{ar^3} = \frac{13122}{54} \\[1em] \Rightarrow r^5 = 243 \\[1em] \Rightarrow r^5 = 3^5 \\[1em] \Rightarrow r = 3.\end{aligned}$$

Substituting r in (i) we get,

$$\begin{aligned}\Rightarrow a(3)^3 = 54 \\[1em] \Rightarrow 27a = 54 \\[1em] \Rightarrow a = 2.\end{aligned}$$

n^th term of a G.P. = ar^{n - 1}

= 2(3)^{n - 1}

= 2 × 3^{n - 1}

2^nd term of a G.P. = 2 × 3^{2 - 1}

= 2 × 3

= 6.

3^rd term of a G.P. = 2 × 3^{3 - 1}

= 2 × 3²

= 18.

G.P. = 2, 6, 18, 54, .........

Hence, G.P. = 2, 6, 18, 54, ......... and general term = 2 × 3^{n - 1}

Let first term of the G.P. be A and it's common ratio be R.

Given,

⇒ a₅ = p

⇒ AR⁴ = p ........(i)

Also,

⇒ a₈ = q

⇒ AR⁷ = q .........(ii)

⇒ a₁₁ = r

⇒ AR¹⁰ = r .........(iii)

Multiplying (i) by (iii) we get,

AR⁴ x AR¹⁰ = pr

⇒ A²R¹⁴ = pr .........(iv)

Squaring eq. (ii) we get,

(AR⁷)² = q²

A²R¹⁴ = q² .........(v)

L.H.S. of eq. (iv) and (v) are equal so, R.H.S. will also be equal.

∴ q² = pr.

Hence, proved that q² = pr.

Since,

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

So, the above sequence is a G.P. with r = $\sqrt{2}$.

Let there be n terms.

$$\begin{aligned}\Rightarrow ar^{n - 1} = 32 \\[1em] \Rightarrow \sqrt{2} \times (\sqrt{2})^{n - 1} = 32 \\[1em] \Rightarrow \sqrt{2}^{n - 1 + 1} = 32 \\[1em] \Rightarrow (\sqrt{2})^n = (\sqrt{2})^{10} \\[1em] \Rightarrow n = 10.\end{aligned}$$

m^th term from end is (n - m + 1)^th term from starting.

So, 7^th term from end is (10 - 7 + 1) = 4^th term from starting.

a₄ = ar³

= $\sqrt{2}(\sqrt{2})^3$

= $\sqrt{2} \times 2\sqrt{2}$

= 4.

Hence, 7^th term from end is 4.

Common ratio,

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

So, the above sequence is a G.P. with r = 3.

Let there be n terms.

$$\begin{aligned}\Rightarrow ar^{n - 1} = 162 \\[1em] \Rightarrow \frac{2}{27} \times (3)^{n - 1} = 162 \\[1em] \Rightarrow \frac{2}{3^3} \times 3^{n - 1} = 162 \\[1em] \Rightarrow 3^{n - 1 - 3} = \frac{162}{2} \\[1em] \Rightarrow 3^{n - 4} = 81 \\[1em] \Rightarrow 3^{n - 4} = 3^4 \\[1em] \Rightarrow n - 4 = 4 \\[1em] \Rightarrow n = 8.\end{aligned}$$

m^th term from end is (n - m + 1)^th term from starting.

So, 3^rd term from end is (8 - 3 + 1) = 6^th term from starting.

a₆ = ar⁵

= $\frac{2}{27} \times (3)^5$

= $\frac{2}{27} \times 243$

= 2 × 9

= 18.

Hence, 3^rd term from end is 18.

Common ratio,

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

So, the above sequence is a G.P. with r = 3.

Let there be n terms.

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

a₄ = ar³

= $\frac{1}{27} \times (3)^3$

= $\frac{1}{27} \times 27$

= 1.

m^th term from end is (n - m + 1)^th term from starting.

So, 4^th term from end is (8 - 4 + 1) = 5^th term from starting.

a₅ = ar⁴

= $\frac{1}{27} \times (3)^4$

= $\frac{1}{27} \times 81$

= 3.

a₄.a₅ = 1 × 3 = 3.

Hence, product of fourth term from the beginning and the fourth term from the end = 3.

Given,

a, b and c are in G.P.

⇒ b² = ac .............(i)

a, x, b, y, c are in A.P.

⇒ 2x = a + b and 2y = b + c.

⇒ $x = \frac{a + b}{2} \text{ and } y = \frac{b + c}{2}$ ........(ii)

(i) Substituting value of x and y from (ii) in L.H.S. of $\frac{1}{x} + \frac{1}{y} = \frac{2}{b}$,

$$\begin{aligned}\text{L.H.S}. = \frac{1}{\frac{a + b}{2}} + \frac{1}{\frac{b + c}{2}} \\[1em] = \frac{2}{a + b} + \frac{2}{b + c} \\[1em] = \frac{2(b + c) + 2(a + b)}{(a + b)(b + c)} \\[1em] = \frac{2b + 2c + 2a + 2b}{(a + b)(b + c)}\\[1em] = \frac{2a + 2c + 4b}{ab + ac + b^2 + bc} \\[1em] = \frac{2(a + c + 2b)}{ab + b^2 + b^2 + bc} \text{from (i)} \\[1em] = \frac{2(a + c + 2b)}{b(a + b + b + c)} \\[1em] = \frac{2(a + c + 2b)}{b(a + c + 2b)} \\[1em] = \frac{2}{b} = \text{R.H.S.}\end{aligned}$$

Hence, proved that $\frac{1}{x} + \frac{1}{y} = \frac{2}{b}$.

(ii) Substituting value of x and y from (ii) in L.H.S. of $\frac{a}{x} + \frac{c}{y} = 2$,

$$\begin{aligned}\text{L.H.S.} = \frac{a}{\frac{a + b}{2}} + \frac{c}{\frac{b + c}{2}} \\[1em] = \frac{2a}{a + b} + \frac{2c}{b + c} \\[1em] = 2\Big(\frac{a}{a + b} + \frac{c}{b + c}\Big) \\[1em] = 2\Big[\frac{a(b + c) + c(a + b)}{(a + b)(b + c)}\Big] \\[1em] = 2\Big(\frac{ab + ac + ac + bc}{ab + ac + b^2 + bc}\Big) \\[1em] = 2\Big(\frac{ab + b^2 + b^2 + bc}{ab + b^2 + b^2 + bc}\Big) ......\text{from (i)} \\[1em] = 2 = \text{R.H.S.}\end{aligned}$$

Hence, proved that $\frac{a}{x} + \frac{c}{y} = 2$.

Since, a, b and c are in A.P.

⇒ 2b = a + c

⇒ b = $\frac{a + c}{2}$ .......(i)

Since, a, b and c are also in G.P.

⇒ b² = ac

Substituting value of b from (i) in above equation we get,

$$\begin{aligned}\Rightarrow \Big(\frac{a + c}{2}\Big)^2 = ac \\[1em] \Rightarrow \frac{a^2 + c^2 + 2ac}{4} = ac \\[1em] \Rightarrow a^2 + c^2 + 2ac = 4ac \\[1em] \Rightarrow a^2 + c^2 - 2ac = 0 \\[1em] \Rightarrow (a - c)^2 = 0\\[1em] \Rightarrow a - c = 0 \\[1em] \Rightarrow a = c .......(ii)\end{aligned}$$

Substituting value of a from (ii) in (i) we get,

b = $\frac{c + c}{2} = \frac{2c}{2}$ = c .......(iii)

From (ii) and (iii) we get,

a = b = c.

Hence, proved that a = b = c.

By formula,

Common ratio = $\frac{a_{n + 1}}{a_n}$

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

By formula,

⇒ a_n = ar^{n - 1}

⇒ a₄ = 2 × (3)^{4 - 1}

⇒ a₄ = 2 × 3³

⇒ a₄ = 2 × 27 = 54.

Sum of first four terms of G.P. = 2 + 6 + 18 + 54 = 80.

Hence, Option 2 is the correct option.

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

By formula,

⇒ a_n = ar^{n - 1}

Given,

4^th term of a G.P. is 54.

⇒ a₄ = 54

⇒ ar^{4 - 1} = 54

⇒ ar³ = 54 .........(1)

7^th term of a G.P. is 1458.

⇒ a₇ = 1458

⇒ ar^{7 - 1} = 1458

⇒ ar⁶ = 1458 .........(2)

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

$$\begin{aligned}\Rightarrow \frac{ar^6}{ar^3} = \frac{1458}{54} \\[1em] \Rightarrow r^3 = 27 \\[1em] \Rightarrow r^3 = 3^3 \\[1em] \Rightarrow r = 3.\end{aligned}$$

Hence, Option 2 is the correct option.

Given,

8, x and 32 are in G.P.

$$\begin{aligned}\Rightarrow \frac{x}{8} = \frac{32}{x} \\[1em] \Rightarrow x^2 = 8 \times 32 \\[1em] \Rightarrow x^2 = 256 \\[1em] \Rightarrow x = \sqrt{256} = 16.\end{aligned}$$

Hence, Option 4 is the correct option.

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

Given,

Product of three terms of G.P. = 1

$$\begin{aligned}\therefore \frac{a}{r} \times a \times ar = 1 \\[1em] \Rightarrow a^3 = 1 \\[1em] \Rightarrow a^3 = 1^3 \\[1em] \Rightarrow a = 1.\end{aligned}$$

Given,

Sum of three terms of G.P. = $3\frac{1}{2}$

$$\begin{aligned}\Rightarrow \frac{a}{r} + a + ar = 3\frac{1}{2} \\[1em] \Rightarrow \frac{1}{r} + 1 + 1(r) = \frac{7}{2} \quad [\because a = 1] \\[1em] \Rightarrow \frac{1}{r} + 1 + r = \frac{7}{2} \\[1em] \Rightarrow \frac{1 + r + r^2}{r} = \frac{7}{2} \\[1em] \Rightarrow 2(r^2 + r + 1) = 7r \\[1em] \Rightarrow 2r^2 + 2r + 2 = 7r \\[1em] \Rightarrow 2r^2 + 2r - 7r + 2 = 0 \\[1em] \Rightarrow 2r^2 - 5r + 2 = 0 \\[1em] \Rightarrow 2r^2 - 4r - r + 2 = 0 \\[1em] \Rightarrow 2r(r - 2) - 1(r - 2) = 0 \\[1em] \Rightarrow (2r - 1)(r - 2) = 0 \\[1em] \Rightarrow 2r - 1 = 0 \text{ or } r - 2 = 0 \\[1em] \Rightarrow 2r = 1 \text{ or } r = 2 \\[1em] \Rightarrow r = \frac{1}{2} \text{ or } r = 2.\end{aligned}$$

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

Terms :

⇒ $\frac{a}{r}$, a, ar

⇒ $\frac{1}{\frac{1}{2}}, 1, 1 \times \frac{1}{2}$

⇒ 2, 1, $\frac{1}{2}$.

Let r = 2

Terms :

⇒ $\frac{a}{r}$, a, ar

⇒ $\frac{1}{2}, 1, 1 \times 2$

⇒ $\frac{1}{2}$, 1, 2.

Hence, Option 1 is the correct option.

By formula,

Common ratio = $\frac{a_{n + 1}}{a_n}$

Given,

G.P. = 10, 20, 40, ......

a = 10 and r = $\frac{20}{10}$ = 2.

We know that,

If | r | > 1

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

Substituting values we get :

$$\begin{aligned}\Rightarrow S_{20} = \frac{10(2^{20} - 1)}{2 - 1} \\[1em] = \frac{10(2^{20} - 1)}{1} \\[1em] = 10(2^{20} - 1).\end{aligned}$$

Hence, Option 3 is the correct option.

Common ratio (r) = $\frac{3}{1}$ = 3.

$$\begin{aligned}S = \frac{a(r^n - 1)}{(r - 1)} ..........(As |r| \gt 1)\\[1em] = \frac{1(3^{12} - 1)}{3 - 1} \\[1em] = \frac{531441 - 1}{2} \\[1em] = \frac{531440}{2} \\[1em] = 265720.\end{aligned}$$

Hence, sum = 265720.

Common ratio (r) = $\frac{0.03}{0.3}$ = 0.1

$$\begin{aligned}S = \frac{a(1 - r^n)}{(1 - r)} ..........(As |r| \lt 1)\\[1em] = \frac{0.3\Big[1 - (0.1)^8\Big]}{1 - 0.1} \\[1em] = \frac{0.3\Big[1 - \Big(\frac{1}{10}\Big)^8\Big]}{0.9} \\[1em] = \frac{1}{3}\Big(1 - \frac{1}{10^8}\Big).\end{aligned}$$

Hence, sum = $\frac{1}{3}\Big(1 - \frac{1}{10^8}\Big)$.

Common ratio (r) = $\frac{-\frac{1}{2}}{1} = -\frac{1}{2}$

$$\begin{aligned}S = \frac{a(1 - r^n)}{(1 - r)} ..........(As |r| \lt 1) \\[1em] = \frac{1\Big[1 - \Big(-\frac{1}{2}\Big)^9\Big]}{1 - \Big(-\frac{1}{2}\Big)} \\[1em] = \frac{\Big[1 +\Big(\frac{1}{2^9}\Big)\Big]}{1 + \frac{1}{2}} \\[1em] = \frac{\Big(1 + \frac{1}{2^9}\Big)}{\frac{3}{2}} \\[1em] = \frac{2}{3}\Big(1 + \frac{1}{2^9}\Big)\end{aligned}$$

Hence, sum = $\frac{2}{3}\Big(1 + \frac{1}{2^9}\Big)$.

Let n terms be added.

Common ratio = $\frac{4}{1}$ = 4.

$$\begin{aligned}\Rightarrow S = \frac{a(r^n - 1)}{(r - 1)} ..........(As |r| \gt 1)\\[1em] \Rightarrow 5461 = \frac{1[(4)^n - 1]}{4 - 1} \\[1em] \Rightarrow \frac{4^n - 1}{3} = 5461 \\[1em] \Rightarrow 4^n - 1 = 16383 \\[1em] \Rightarrow 4^n = 16384 \\[1em] \Rightarrow 4^n = 4^7 \\[1em] \Rightarrow n = 7.\end{aligned}$$

Hence, 7 terms must be added to get a sum of 5461.

Given , a = 27 and a₈ = $\frac{1}{81}$.

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

Since, r < 1

$$\begin{aligned}S = \frac{a(1 - r^n)}{(1 - r)} \\[1em] = \frac{27\Big[1 - \Big(\frac{1}{3}\Big)^{10}\Big]}{1 - \frac{1}{3}} \\[1em] = \frac{27\Big(1 - \frac{1}{3^{10}}\Big)}{\frac{2}{3}} \\[1em] = \frac{81}{2}\Big(1 - \frac{1}{3^{10}}\Big).\end{aligned}$$

Hence, sum upto 10 terms = $\frac{81}{2}\Big(1 - \frac{1}{3^{10}}\Big).$

G.P. formed = 10 + 20 + 40 + ........

Common ratio (r) = $\frac{20}{10}$ = 2.

$$\begin{aligned}\Rightarrow S = \frac{a(r^n - 1)}{(r - 1)} ..........(As |r| \gt 1)\\[1em] = \frac{10(2^{12} - 1)}{2 - 1} \\[1em] = 10(2^{12} - 1).\end{aligned}$$

Hence, ₹10(2¹² - 1).

Given,

$a_4 = \frac{1}{27}$

$ar^3 = \frac{1}{27}$ .........(i)

$a_7 = \frac{1}{729}$

$ar^6 = \frac{1}{729}$ ........(ii)

Dividing (ii) by (i) we get,

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

Substituting value of r in (i) we get,

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

Since, r < 1

$$\begin{aligned}S = \frac{a(1 - r^n)}{(1 - r)} \\[1em] = \frac{1\Big[1 - \Big(\frac{1}{3}\Big)^n \Big]}{1 - \frac{1}{3}} \\[1em] = \frac{\Big[1 - \Big(\frac{1}{3}\Big)^n \Big]}{\frac{2}{3}} \\[1em] = \frac{3}{2}\Big[1 - \Big(\frac{1}{3}\Big)^n \Big] \\[1em] = \frac{3}{2}\Big(1 - \frac{1}{3^n} \Big)\end{aligned}$$

Hence, sum = $\frac{3}{2}\Big(1 - \frac{1}{3^n} \Big).$

Let n^th term be the last term.

⇒ ar^{n - 1} = 486

⇒ a(3)^{n - 1} = 486

⇒ a.3ⁿ.3^-1 = 486

⇒ $3^n = \frac{486 \times 3}{a} = \frac{1458}{a}$

Since, r > 1

$$\begin{aligned}\Rightarrow S = \frac{a(r^n - 1)}{(r - 1)} \\[1em] \Rightarrow 728 = \frac{a \times (3^n - 1)}{3 - 1} \\[1em] \Rightarrow 728 = \frac{a \times \Big(\frac{1458}{a} - 1\Big)}{2} \\[1em] \Rightarrow 728 = \frac{a \times (1458 - a)}{2a} \\[1em] \Rightarrow \frac{1458 - a}{2} = 728 \\[1em] \Rightarrow 1458 - a = 1456 \\[1em] \Rightarrow a = 1458 - 1456 \\[1em] \Rightarrow a = 2.\end{aligned}$$

Hence, first term = 2.

Common ratio = $\frac{6}{3}$ = 2.

Let 1536 be n^th term

$$\begin{aligned}\therefore ar^{n - 1} = 1536 \\[1em] \Rightarrow 3.(2)^{n - 1} = 1536 \\[1em] \Rightarrow (2)^{n - 1} = 512 \\[1em] \Rightarrow (2)^{n - 1} = (2)^{9} \\[1em] \Rightarrow n - 1 = 9 \\[1em] \Rightarrow n = 10.\end{aligned}$$

Since, r > 1

$$\begin{aligned}S = \frac{a(r^n - 1)}{(r - 1)} \\[1em] = \frac{3 \times (2^{10} - 1)}{2 - 1} \\[1em] = 3(2^{10} - 1) \\[1em] = 3(1024 - 1) \\[1em] = 3 \times 1023 \\[1em] = 3069.\end{aligned}$$

Hence, sum = 3069.

Common ratio = $\frac{6}{2}$ = 3.

Let n be no. of terms taken.

Since, r > 1

$$\begin{aligned}\Rightarrow S = \frac{a(r^n - 1)}{(r - 1)} \\[1em] \Rightarrow 728 = \frac{2 \times (3^n - 1)}{3 - 1} \\[1em] \Rightarrow 728 = \frac{2(3^n - 1)}{2} \\[1em] \Rightarrow 3^n - 1 = 728 \\[1em] \Rightarrow 3^n = 729 \\[1em] \Rightarrow 3^n = 3^6 \\[1em] \Rightarrow n = 6.\end{aligned}$$

Hence, 6 terms must be taken to make the sum equal to 728.

Given,

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

Hence, common ratio = $\frac{3}{5}$.

Common ratio = 2.

Since, |r| > 1

$$\begin{aligned}\Rightarrow S = \frac{a(r^n - 1)}{(r - 1)} \\[1em] \Rightarrow 255 = \frac{1(2^n - 1)}{(2 - 1)} \\[1em] \Rightarrow 2^n - 1 = 255 \\[1em] \Rightarrow 2^n = 256 \\[1em] \Rightarrow 2^n = 2^8 \\[1em] \Rightarrow n = 8.\end{aligned}$$

Hence, n = 8.

Let the numbers be $\frac{a}{r}, a, ar$.

Given,

Product = 1.

$$\begin{aligned}\therefore \frac{a}{r} \times a \times ar = 1 \\[1em] \Rightarrow a^3 = 1 \\[1em] \Rightarrow a = 1.\end{aligned}$$

Sum = $\frac{39}{10}$

$$\begin{aligned}\therefore \frac{a}{r} + a + ar = \frac{39}{10} \\[1em] \Rightarrow \frac{1}{r} + 1 + r = \frac{39}{10} \\[1em] \Rightarrow \frac{1 + r + r^2}{r} = \frac{39}{10} \\[1em] \Rightarrow 10(1 + r + r^2) = 39r \\[1em] \Rightarrow 10 + 10r + 10r^2 = 39r \\[1em] \Rightarrow 10r^2 - 29r + 10 = 0 \\[1em] \Rightarrow 10r^2 - 25r - 4r + 10 = 0 \\[1em] \Rightarrow 5r(2r - 5) - 2(2r - 5) = 0 \\[1em] \Rightarrow (5r - 2)(2r - 5) = 0 \\[1em] \Rightarrow 5r - 2 = 0 \text{ or } 2r - 5 = 0 \\[1em] \Rightarrow r = \frac{2}{5} \text{ or } r = \frac{5}{2}\end{aligned}$$

Let r = $\frac{2}{5}$

Numbers = $\frac{a}{r} = \frac{1}{\frac{2}{5}} = \frac{5}{2}$

a = 1

ar = $1 \times \frac{2}{5} =\frac{2}{5}$.

Let r = $\frac{5}{2}$

Numbers = $\frac{a}{r} = \frac{1}{\frac{5}{2}} = \frac{2}{5}$

a = 1

ar = $1 \times \frac{5}{2} =\frac{5}{2}$.

Hence, numbers = $\frac{5}{2}, 1, \frac{2}{5} \text{ or } \frac{2}{5}, 1, \frac{5}{2}$.

Given, a = -3 and,

(a₂)² = a₄

(ar)² = ar³

(-3r)² = -3r³

9r² = -3r³

r = -3.

a₇ = ar⁶ = (-3)(-3)⁶

= -3 × 729

= -2187.

Hence, 7^th term = -2187.

Common ratio = $\frac{1}{\frac{5}{2}} = \frac{2}{5}$

a₅ = ar⁴

= $\frac{5}{2} \times \Big(\frac{2}{5}\Big)^4$

= $\frac{5}{2} \times \frac{16}{625}$

= $\frac{8}{125}$.

Hence, 5^th term = $\frac{8}{125}$.

Given,

⇒ a = 125

⇒ a₂ = 25

⇒ ar = 25

⇒ 125r = 25

⇒ r = $\frac{25}{125} = \frac{1}{5}$.

a₅ = ar⁴

= $125 \times \Big(\frac{1}{5}\Big)^4$

= $125 \times \frac{1}{625}$

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

a₆ = ar⁵

= $125 \times \Big(\frac{1}{5}\Big)^5$

= $125 \times \frac{1}{3125}$

= $\frac{1}{25}.$

Hence, $a_5 = \frac{1}{5} \text{ and } a_6 = \frac{1}{25}.$

Common ratio = $\frac{1}{-\frac{1}{3}} = -3$.

Since, |r| > 1

$$\begin{aligned}S = \frac{a(r^n - 1)}{(r - 1)} \\[1em] = \frac{-\frac{1}{3}[(-3)^8 - 1]}{-3 - 1} \\[1em] = -\frac{1}{3} \times \frac{[(-1)^8(3)^8 - 1]}{-4} \\[1em] = \frac{1}{12}(3^8 - 1).\end{aligned}$$

Hence, sum = $\frac{1}{12}(3^8 - 1).$

Given,

⇒ a = 27

⇒ a₈ = ar⁷ = $\frac{1}{81}$

⇒ 27r⁷ = $\frac{1}{81}$

⇒ r⁷ = $\frac{1}{81 \times 27}$

⇒ r⁷ = $\frac{1}{3^7}$

⇒ r⁷ = $\Big(\frac{1}{3}\Big)^7$

⇒ r = $\frac{1}{3}$.

Since, r < 1

$$\begin{aligned}S = \frac{a(1 - r^n)}{(1 - r)} \\[1em] = \frac{27\Big[1 - \Big(\frac{1}{3}\Big)^{10}\Big]}{1 - \frac{1}{3}} \\[1em] = \frac{27\Big[1 - \Big(\frac{1}{3}\Big)^{10}\Big]}{ \frac{2}{3}} \\[1em] = \frac{81}{2}\Big[1 - \Big(\frac{1}{3}\Big)^{10}\Big] \\[1em] = \frac{81}{2}(1 - 3^{-10}).\end{aligned}$$

Hence, sum = $\frac{81}{2}(1 - 3^{-10}).$

Given,

⇒ a + a₂ = -4

⇒ a + ar = -4

⇒ a(1 + r) = -4 ........(i)

⇒ a₅ = 4a₃

⇒ ar⁴ = 4ar²

⇒ r² = 4

⇒ r = ± 2

Let r = 2.

Substituting value of r in (i) we get,

⇒ a(1 + 2) = -4

⇒ 3a = -4

⇒ a = $-\frac{4}{3}$.

G.P. = a, ar, ar², ........

= $-\frac{4}{3}, -\frac{8}{3}, -\frac{16}{3}$........

Let r = -2.

Substituting value of r in (i) we get,

⇒ a(1 + (-2)) = -4

⇒ -a = -4

⇒ a = 4.

G.P. = a, ar, ar², ........

= 4, -8, 16, -32,........

Hence, G.P. = $-\frac{4}{3}, -\frac{8}{3}, -\frac{16}{3}$........ or 4, -8, 16, -32.

Given,

-1, k and -1 are three consecutive terms of a G.P.

$$\begin{aligned}\therefore \frac{k}{-1} = \frac{-1}{k} \\[1em] \Rightarrow k \times k = -1 \times -1 \\[1em] \Rightarrow k^2 = 1 \\[1em] \Rightarrow k = \sqrt{1} = \pm 1.\end{aligned}$$

Hence, Option 4 is the correct option.

Given,

x + 9, 10 and 4 are in G.P.

$$\begin{aligned}\therefore \frac{10}{x + 9} = \frac{4}{10} \\[1em] \Rightarrow 4(x + 9) = 10 \times 10 \\[1em] \Rightarrow 4x + 36 = 100 \\[1em] \Rightarrow 4x = 100 - 36 \\[1em] \Rightarrow 4x = 64 \\[1em] \Rightarrow x = \frac{64}{4} = 16.\end{aligned}$$

Hence, Option 1 is the correct option.

Given,

Common ratio of G.P. (r) = 2

By formula,

⇒ a_n = ar^{n - 1}

Given,

6^th term of G.P. is 48.

⇒ a₆ = 48

⇒ a × (2)^{6 - 1} = 48

⇒ a × 2⁵ = 48

⇒ 32a = 48

⇒ a = $\frac{48}{32} = \frac{3}{2}$.

Hence, Option 1 is the correct option.

Let three terms be $\frac{a}{r}$, a and ar.

Given,

Product of three terms = 27

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

Hence, Option 2 is the correct option.

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

By formula,

⇒ a_n = ar^{n - 1}

Given,

4^th term of G.P. is 27.

⇒ a₄ = 27

⇒ ar^{4 - 1} = 27

⇒ ar³ = 27 ........(1)

Given,

6^th term of G.P. is 243.

⇒ a₆ = 243

⇒ ar^{6 - 1} = 243

⇒ ar⁵ = 243 ........(2)

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

$$\begin{aligned}\Rightarrow \frac{ar^5}{ar^3} = \frac{243}{27} \\[1em] \Rightarrow r^2 = 9 \\[1em] \Rightarrow r = \sqrt{9} = 3.\end{aligned}$$

Hence, Option 2 is the correct option.

Let first five terms of G.P. be

$\frac{a}{r^2}, \frac{a}{r}, a, ar, ar^2$

Given,

Third term of a G.P. = 18.

∴ a = 18

Product of first five terms = $\frac{a}{r^2} \times \frac{a}{r} \times a \times ar \times ar^2$

= a⁵

= 18⁵.

Hence, Option 2 is the correct option.

Given, the sequence = $\frac{2}{9}, \frac{1}{3}, \frac{1}{2},..............$

First term (a) = $\frac{2}{9}$

Common ratio (r) = $\frac{\frac{1}{3}}{\frac{2}{9}} = \frac{1 \times 9}{2 \times 3} = \frac{3}{2}$

Using the formula; T_n = a.r^{n - 1}

$$\begin{aligned}T_5 = \frac{2}{9} × \Big(\frac{3}{2}\Big)^{5 - 1}\\[1em] = \frac{2}{9} × \Big(\frac{3}{2}\Big)^4\\[1em] = \frac{2}{9} × \frac{81}{16}\\[1em] = \frac{9}{8}\\[1em] = 1\frac{1}{8}.\end{aligned}$$

So, assertion (A) is true.

According to reason :

Sum of first n terms (S_n) = $\frac{a(r^n - 1)}{r - 1}$, for all r.

But this is not the case when r = 1.

Thus,

Reason (R) is false.

Hence, option 1 is the correct option.

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

By formula :

⇒ T_n = a.r^{n - 1}

Given, fourth term = x, seventh term = y and tenth term = z

⇒ a₄ = x, a₇ = y and a₁₀ = z

⇒ ar^{4 - 1} = x, ar^{7 - 1} = y and ar^{10 - 1} = z

⇒ ar³ = x, ar⁶ = y and ar⁹ = z

If x, y and z are in G.P., then the ratio between the consecutive terms will be equal.

Ratio between y and x :

$\Rightarrow \frac{y}{x} = \frac{ar^6}{ar^3} = r^3.$

Ratio between z and y :

$\Rightarrow \frac{z}{y} = \frac{ar^9}{ar^6} = r^3.$

Since, the ratio between the consecutive terms are equal.

Thus, x, y and z are in G.P.

∴ Assertion (A) is true.

⇒ y² = (ar⁶)²

⇒ y² = ar¹²

⇒ y² = ar³ × ar⁹

⇒ y² = xz.

∴ Reason (R) is true.

Hence, option 3 is the correct option.

Given, G.P.: = 3 - 6 + 12 - 24 + ............. - 384

Here, a = 3

common ratio, r = $\frac{-6}{3}$ = -2

a_n = -384

Using the formula; T_n = a.r^{n - 1}

⇒ 3 x (-2)^{n - 1} = -384

⇒ (-2)^{n - 1} = $-\frac{384}{3}$

⇒ (-2)^{n - 1} = -128

⇒ (-2)^{n - 1} = (-2)⁷

⇒ n - 1 = 7

⇒ n = 7 + 1 = 8

5_th term from the beginning,

⇒ T₅ = 3 x (-2)^{5 - 1}

= 3 x (-2)⁴

= 3 x 16 = 48

5_th term from the last = (8 - 5 + 1) = 4_th term from the beginning,

⇒ T₄ = 3 x (-2)^{4 - 1}

= 3 x (-2)³

= 3 x (-8) = -24

Product of 5_thterm from the beginning and 5_thterm from the end = 48 x (-24) = -1152.

So, statement 1 is true.

Let in a G.P.

a be the first term and N be the total number of terms.

n^th term from the beginning,

⇒ T_n = a.r^{n - 1} ........(1)

n^th term from the end,

⇒ T_{N - n + 1} = a.r^{N - n + 1 - 1}

⇒ T_{N - n + 1} = ar^{N - n} .....(2)

Multiplying equation (1) and (2), we get :

⇒ T_n x T_{N - n + 1} = a.r^{n - 1} x a.r^{N - n}

= a².r^{(n - 1) + (N - n)}

= a².r^{N - 1} ....................(3)

Now, first term + last term = a + a.r^{n - 1}

= a(1 + r^{n - 1}) ....................(4)

From equation (3) and (4),

The product of n^th term from the beginning and n^th term from the end is not equal to 1^st term + last term.

So, statement 2 is false.

Hence, option 3 is the correct option.

Given,

First term, a = 3

Common ratio, r = 2

Last term = 96

Using the formula;

⇒ T_n = a.r^{n - 1}

⇒ 3 x 2^{n - 1} = 96

⇒ 2^{n - 1} = $\frac{96}{3}$

⇒ 2^{n - 1} = 32

⇒ 2^{n - 1} = 2⁵

⇒ n - 1 = 5

⇒ n = 5 + 1 = 6

And, according to statement 1, the number of terms in this G.P. = 96 - 3 = 93, which is not correct.

So, statement 1 is false.

$$\begin{aligned}\Rightarrow \frac{a}{ar^{n - 1}} = \frac{\text{first term}}{\text{last term}}\\[1em] = \frac{3}{96} = 3 : 96.\end{aligned}$$

So, statement 2 is true.

Hence, option 4 is the correct option.

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

By formula,

⇒ a_n = ar^{n - 1}

Given,

5^th term of G.P. = 32

⇒ a₅ = 32

⇒ ar^{5 - 1} = 32

⇒ ar⁴ = 32 ..........(1)

Given,

8^th term of G.P. = 256

⇒ a₈ = 256

⇒ ar^{8 - 1} = 256

⇒ ar⁷ = 256 ..........(2)

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

$$\begin{aligned}\Rightarrow \frac{ar^7}{ar^4} = \frac{256}{32} \\[1em] \Rightarrow r^3 = 8 \\[1em] \Rightarrow r^3 = 2^3 \\[1em] \Rightarrow r = 2.\end{aligned}$$

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

⇒ a × (2)⁴ = 32

⇒ 16a = 32

⇒ a = $\frac{32}{16}$ = 2.

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

Let first term be a and common ratio be r.

By formula,

⇒ a_n = ar^{n - 1}

Given,

The third term of a G.P. is greater than its first term by 9.

∴ a₃ - a = 9

⇒ ar^{3 - 1} - a = 9

⇒ ar² - a = 9

⇒ a(r² - 1) = 9

⇒ a = $\frac{9}{r^2 - 1}$ .......(1)

Given,

Second term is greater than the fourth term by 18.

∴ a₂ - a₄ = 18

⇒ ar^{2 - 1} - ar^{4 - 1} = 18

⇒ ar - ar³ = 18

⇒ ar(1 - r²) = 18

Substituting value of a from equation (1) in above equation, we get :

$$\begin{aligned}\Rightarrow \frac{9}{r^2 - 1} \times r(1 - r^2) = 18 \\[1em] \Rightarrow \frac{9}{r^2 - 1} \times -r(r^2 - 1) = 18 \\[1em] \Rightarrow -9r = 18 \\[1em] \Rightarrow r = -\frac{18}{9} = -2.\end{aligned}$$

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

$$\begin{aligned}\Rightarrow a = \frac{9}{(-2)^2 - 1} \\[1em] = \frac{9}{4 - 1} \\[1em] = \frac{9}{3} \\[1em] = 3.\end{aligned}$$

G.P. = a, ar, ar², ar³, .......

= 3, 3 × -2, 3 × (-2)², 3 × (-2)³, .......

= 3, -6, 3 × 4, 3 × -8, ........

= 3, -6, 12, -24, ........

Hence, required G.P. = 3, -6, 12, -24, ........

Given,

x, 2x + 2, 3x + 3, G are four consecutive terms of a G.P.

$$\begin{aligned}\therefore \frac{2x + 2}{x} = \frac{3x + 3}{2x + 2} \\[1em] \Rightarrow (2x + 2)^2 = x(3x + 3) \\[1em] \Rightarrow (2x)^2 + 2^2 + 2 \times 2x \times 2 = 3x^2 + 3x \\[1em] \Rightarrow 4x^2 + 4 + 8x = 3x^2 + 3x \\[1em] \Rightarrow 4x^2 - 3x^2 + 8x - 3x + 4 = 0 \\[1em] \Rightarrow x^2 + 5x + 4 = 0 \\[1em] \Rightarrow x^2 + 4x + x + 4 = 0 \\[1em] \Rightarrow x(x + 4) + 1(x + 4) = 0 \\[1em] \Rightarrow (x + 1)(x + 4) = 0 \\[1em] \Rightarrow x + 1 = 0 \text{ or } x + 4 = 0 \\[1em] \Rightarrow x = -1 \text{ or } x = -4.\end{aligned}$$

Substituting value of x = -1, in terms we get :

Terms : -1, 2(-1) + 2, 3(-1) + 3, G

= -1, -2 + 2, -3 + 3, G

= -1, 0, 0, G

This is not possible as in this case common ratio is different.

Substituting value of x = -4, in terms we get :

Terms : -4, 2(-4) + 2, 3(-4) + 3, G

= -4, -8 + 2, -12 + 3, G

= -4, -6, -9, G

Here, common difference = $\frac{-6}{-4} = \frac{3}{2}$.

G = $-9 \times \frac{3}{2} = -\frac{27}{2}$.

Hence, G = $-\frac{27}{2}$.

Let first five terms of G.P. be $\frac{a}{r^2}, \frac{a}{r}, a, ar, ar^2$.

Given,

Third term of the G.P. is 2.

∴ a = 2

Product of terms = $\frac{a}{r^2} \times \frac{a}{r} \times a \times ar \times ar^2 = a^5$

= 2⁵

= 32.

Hence, product of the first five terms of this G.P. = 32.

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

By formula,

⇒ a_n = ar^{n - 1}

Given,

The 10^th, 16^th and 22^nd terms of a G.P. are x, y and z respectively.

⇒ a₁₀ = x

⇒ x = ar^{10 - 1}

⇒ x = ar⁹ ...........(1)

⇒ a₁₆ = y

⇒ y = ar^{16 - 1}

⇒ y = ar¹⁵ ...........(2)

⇒ a₂₂ = z

⇒ z = ar^{22 - 1}

⇒ z = ar²¹ ...........(3)

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

$$\begin{aligned}\Rightarrow \frac{y}{x} = \frac{ar^{15}}{ar^9} \\[1em] \Rightarrow \frac{y}{x} = \frac{r^{15}}{r^9} \\[1em] \Rightarrow \frac{y}{x} = r^{15 - 9} \\[1em] \Rightarrow \frac{y}{x} = r^6.\end{aligned}$$

Dividing equation (3) by (2), we get :

$$\begin{aligned}\Rightarrow \frac{z}{y} = \frac{ar^{21}}{ar^{15}} \\[1em] \Rightarrow \frac{z}{y} = \frac{r^{21}}{r^{15}} \\[1em] \Rightarrow \frac{z}{y} = r^{21 - 15} \\[1em] \Rightarrow \frac{z}{y} = r^6.\end{aligned}$$

Since, $\frac{y}{x} = \frac{z}{y}$ = r⁶.

Hence, proved that x, y and z are in G.P.

Given,

G.P. : $2, 2\sqrt{2}, 4, ........$

In above G.P.,

First term (a) = 2

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

Let n^th term of G.P. be $128\sqrt{2}$.

$$\begin{aligned}\Rightarrow ar^{n-1} = 128\sqrt{2} \\[1em] \Rightarrow 2 \times (\sqrt{2})^{n - 1} = 128\sqrt{2} \\[1em] \Rightarrow 2 \times (2^{\frac{1}{2}})^{n - 1} = (2)^7.\sqrt{2} \\[1em] \Rightarrow 2 \times (2)^{\frac{n - 1}{2}} = 2^7.2^{\frac{1}{2}} \\[1em] \Rightarrow (2)^{1 + \frac{n - 1}{2}} = (2)^{7 + \frac{1}{2}} \\[1em] \Rightarrow (2)^{\frac{n - 1 + 2}{2}} = (2)^{\frac{14 + 1}{2}} \\[1em] \Rightarrow (2)^{\frac{n + 1}{2}} = (2)^{\frac{15}{2}} \\[1em] \Rightarrow \frac{n + 1}{2} = \frac{15}{2} \\[1em] \Rightarrow n + 1 = 15 \\[1em] \Rightarrow n = 15 - 1 = 14.\end{aligned}$$

Hence, 14^th term of G.P. = $128\sqrt{2}$.

Given,

Common ratio (r) = 2

By formula,

⇒ a_n = ar^{n - 1}

Given,

10^th term of G.P. = 768

⇒ a₁₀ = 768

⇒ ar^{10 - 1} = 768

⇒ ar⁹ = 768

⇒ a × 2⁹ = 768

⇒ a × 512 = 768

⇒ a = $\frac{768}{512} = \frac{3}{2}$.

8th term of G.P. is given by :

⇒ a₈ = ar^{8 - 1}

⇒ a₈ = ar⁷

⇒ a₈ = $\frac{3}{2} \times 2^7$

⇒ a₈ = 3 × 2⁶

⇒ a₈ = 3 × 64 = 192.

Hence, 8^th term of a G.P. is given by 192.

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

By formula,

⇒ a_n = ar^{n - 1}

Given,

4^th term of G.P. = 48

⇒ a₄ = 48

⇒ ar^{4 - 1} = 48

⇒ ar³ = 48 ..........(1)

Given,

7^th term of G.P. = 384

⇒ a₇ = 48

⇒ ar^{7 - 1} = 384

⇒ ar⁶ = 384 ..........(2)

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

$$\begin{aligned}\Rightarrow \frac{ar^6}{ar^3} = \frac{384}{48} \\[1em] \Rightarrow r^3 = 8 \\[1em] \Rightarrow r^3 = 2^3 \\[1em] \Rightarrow r = 2.\end{aligned}$$

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

⇒ a(2)³ = 48

⇒ a × 8 = 48

⇒ a = $\frac{48}{8}$ = 6.

⇒ a₆ = ar^{6 - 1}

⇒ a₆ = ar⁵

⇒ a₆ = 6 × 2⁵

⇒ a₆ = 6 × 32

⇒ a₆ = 192.

Hence, 6^th term of G.P. = 192.

Given,

$-\frac{5}{3}, x \text{ and } -\frac{3}{5}$ are three consecutive terms of a G.P.

$$\begin{aligned}\therefore \frac{x}{-\frac{5}{3}} = \frac{-\frac{3}{5}}{x} \\[1em] \Rightarrow x^2= -\frac{3}{5} \times -\frac{5}{3} \\[1em] \Rightarrow x^2 = 1 \\[1em] \Rightarrow x = \sqrt{1} = \pm 1.\end{aligned}$$

Hence, x = +1 or -1.

Rohit writes a letter to four of his friends, and each of them copies the letter and mail it to four different persons and the chain continues. Thus, the above sequence is :

4, 16, 64, .......

The above sequence is a G.P., with first term (a) = 4 and r = 4.

By formula,

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

Thus,

$$\begin{aligned}S_4 = \frac{4 \times (4^4 - 1)}{4 - 1} \\[1em] = \frac{4 \times (256 - 1)}{3} \\[1em] = \frac{4 \times 255}{3} \\[1em] = 4 \times 85 \\[1em] = 340.\end{aligned}$$

Thus, total 340 letters are mailed till fourth set.

If, it costs ₹ 10 to mail one letter.

Total cost = No. of letters × 10 = 340 × ₹ 10 = ₹ 3,400.

Hence, amount spent on postage = ₹ 3,400.

By formula,

a_n = ar^{n - 1}

Given,

The fourth term of a G.P. is eight times its seventh term.

a₄ = 8a₇

ar^{4 - 1} = 8ar^{7 - 1}

ar³ = 8ar⁶

$\frac{r^6}{r^3} = \frac{a}{8a}$

$r^3 = \frac{1}{8}$

r = $\sqrt[3]{\frac{1}{8}} = \frac{1}{2}$.

Given,

Fifth term = $\frac{3}{16}$

$$\begin{aligned}\Rightarrow ar^4 = \frac{3}{16} \\[1em] \Rightarrow a \times \Big(\frac{1}{2}\Big)^4 = \frac{3}{16} \\[1em] \Rightarrow a \times \frac{1}{16} = \frac{3}{16} \\[1em] \Rightarrow a = 3.\end{aligned}$$

12th term :

a₁₂ = ar^{12 - 1}

= ar¹¹

= $3 \times \Big(\frac{1}{2}\Big)^{11}$.

Hence, 12^th term = $3 \times \Big(\frac{1}{2}\Big)^{11}$.

Sequence :

10 + 20 + 40 + ......... upto 10 terms.

The above is a G.P. with first term (a) = 10 and common ratio (r) = 2.

By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow S_{10} = \frac{10 \times (2^{10} - 1)}{2 - 1} \\[1em] = 10 \times (1024 - 1) \\[1em] = 10 \times 1023 \\[1em] = 10,230.\end{aligned}$$

Since, Ramesh lent ₹ 10,000 but received ₹ 10,230 thus he is in profit of ₹ 230.

Hence, Ramesh makes a profit of ₹ 230.

(i) There are 27 flags in total. The middle-most flag is the 14th flag.

This means there are 13 flags to the left and 13 flags to the right of the starting point.

The first flag on either side is 3 meters away. Rohan travels 3 m to place it and 3 m to return.

The second flag is 6 m away. Rohan travels 6 m to place it and 6 m to return.

The first 4 terms (round-trip distances) are :

1st flag = 2 × 3 = 6 m

2nd flag = 2 × 6 = 12 m

3rd flag = 2 × 9 = 18 m

4th flag = 2 × 12 = 24 m

The above terms form an A.P. with common difference = 6.

Hence, the required sequence is an A.P.

(ii) By formula,

a_n = a + (n - 1)d

a₁₀ = 6 + 6(10 - 1)

= 6 + 6 × 9

= 6 + 54

= 60 m.

Hence, the distance covered by Rohan in placing the 10^th flag and returning back to the middle most flag = 60 m.

(iii) By formula,

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

On each side 13 flags are placed.

$$\begin{aligned}S_{13} = \frac{13}{2}[2 \times 6 + (13 - 1) \times 6] \\[1em] = \frac{13}{2} \times [12 + 12 \times 6] \\[1em] = \frac{13}{2} \times [12 + 72] \\[1em] = \frac{13}{2} \times 84 \\[1em] = 13 \times 42 \\[1em] = 546 \text{ m}.\end{aligned}$$

Thus, distance covered on one side = 546 m.

Total distance = 2 × 546 = 1092 m.

Hence, total distance travelled by Rohan in fixing all the 27 flags = 1092 m.