Geometric Progression — Multiple Choice Questions

Which of the following is not a geometric progression ?

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

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

3. -2, 4, -8, 16

4. 2, 0, 4, 0, 8, 0

Answer

In the series,

2, 0, 4, 0, 8, 0

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

∴ 2, 0, 4, 0, 8, 0 is not a G.P.

Hence, Option 4 is the correct option.

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 :

1. lr^{n - 1}

2. lr^{1 - n}

3. $\dfrac{l}{r^{1 - n}}$

4. $\dfrac{l}{r^{n + 1}}$

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.

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

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

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

3. n²a

4. 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.

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

1. (-1)^{(n - 1)} × 2^{(n - 3)}

2. (-1)^{(n - 1)} × 2^{(n - 2)}

3. (-1)^{(n - 1)} × (-2)^{(n - 1)}

4. (-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.

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

1. 1024

2. 2048

3. 4096

4. 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.

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

1. 7th

2. 6th

3. 9th

4. 8th

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.

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 :

1. $\dfrac{a_k}{a_{1}}$

2. $\dfrac{a_1}{a_2}$

3. $\dfrac{a_k}{a_{n-3}}$

4. $\dfrac{a_{k-1}}{a_{k-2}}$

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.

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

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

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

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

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

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.

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

1. $\dfrac{1}{a^2x^2}$

2. $\dfrac{1}{a^4x^4}$

3. a²x²

4. 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.

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

1. 0.1

2. 0.01

3. 1

4. 0.001

Answer

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

= 0.1

Hence, option 1 is the correct option.

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

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

2. x^{(2n + 3)}

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

4. 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.

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

1. a⁹

2. −a¹⁰

3. −a¹¹

4. −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.

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

1. 3 and 2

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

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

4. $\dfrac{3}{2}$ 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.

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

1. 9th

2. 10th

3. 8th

4. 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.

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

1. 256

2. 1024

3. 512

4. 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.

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

1. mn

2. $\sqrt{mn}$

3. (mn)²

4. $\dfrac{m}{n}$

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.

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 :

1. 2

2. 3

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

4. $\dfrac{1}{3}$

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.

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

1. $\dfrac{1}{32}$

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

3. $\dfrac{1}{64}$

4. $\dfrac{1}{128}$

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.

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

1. 18

2. 2

3. 6

4. $\dfrac{2}{3}$

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.

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 :

1. $\dfrac{12}{13}$

2. $\dfrac{11}{13}$

3. $\dfrac{11}{12}$

4. $\dfrac{13}{12}$

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.

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

1. 0, 1

2. 0, −1

3. −1, 1

4. −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.

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

1. 6

2. 9

3. 7

4. 8

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.

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

1. 181

2. 241

3. 381

4. 421

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.

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 :

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

2. $\dfrac{3}{4}$ or $\dfrac{1}{4}$

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

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.

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

1. y = x + z

2. y = xz

3. 2y = x + z

4. $y = \sqrt{xz}$

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.

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

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

2. ar⁴

3. a⁵r¹⁰

4. $\dfrac{a(r^5 - 1)}{r}$

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.

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

1. y² = x²z²

2. y² = x² + z²

3. y² = xz

4. 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.

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

1. $\dfrac{l}{r^{k-1}}$

2. $\dfrac{l}{k}$

3. $\dfrac{al}{r^k}$

4. $\dfrac{l}{ar^{k-1}}$

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.