Geometric Progression — Exercise 11(B)

The sum of first four terms of the G.P. 2, 6, 18, ........., is :

1. 26

2. 80

3. 160

4. 52

Answer

By formula,

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

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

The 4^th term of a G.P. is 54 and its 7^th term is 1458, the common ratio of this G.P. is :

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

2. 3

3. -3

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

Answer

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 :

$\Rightarrow \dfrac{ar^6}{ar^3} = \dfrac{1458}{54} \\[1em] \Rightarrow r^3 = 27 \\[1em] \Rightarrow r^3 = 3^3 \\[1em] \Rightarrow r = 3.$

Hence, Option 2 is the correct option.

8, x and 32 are in G.P., then the value of x is :

1. 24

2. 256

3. 40

4. 16

Answer

Given,

8, x and 32 are in G.P.

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

Hence, Option 4 is the correct option.

The sum of three terms (numbers) of a G.P. is $3\dfrac{1}{2}$ and their product is 1; the numbers are :

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

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

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

4. $2, \dfrac{1}{2}$ and 1

Answer

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

Given,

Product of three terms of G.P. = 1

$\therefore \dfrac{a}{r} \times a \times ar = 1 \\[1em] \Rightarrow a^3 = 1 \\[1em] \Rightarrow a^3 = 1^3 \\[1em] \Rightarrow a = 1.$

Given,

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

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

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

Terms :

⇒ $\dfrac{a}{r}$, a, ar

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

⇒ 2, 1, $\dfrac{1}{2}$.

Let r = 2

Terms :

⇒ $\dfrac{a}{r}$, a, ar

⇒ $\dfrac{1}{2}, 1, 1 \times 2$

⇒ $\dfrac{1}{2}$, 1, 2.

Hence, Option 1 is the correct option.

The sum of 20 terms of the G.P. 10, 20, 40, ...... is :

1. 10(2¹⁹ - 1)

2. 10(2²¹ - 1)

3. 10(2²⁰ - 1)

4. none of these

Answer

By formula,

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

Given,

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

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

We know that,

If | r | > 1

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

Substituting values we get :

$\Rightarrow S_{20} = \dfrac{10(2^{20} - 1)}{2 - 1} \\[1em] = \dfrac{10(2^{20} - 1)}{1} \\[1em] = 10(2^{20} - 1).$

Hence, Option 3 is the correct option.

Find the sum of G.P. :

1 + 3 + 9 + 27 + ........ to 12 terms

Answer

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

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

Hence, sum = 265720.

Find the sum of G.P. :

0.3 + 0.03 + 0.003 + 0.0003 + ...... to 8 terms.

Answer

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

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

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

Find the sum of G.P. :

$1 - \dfrac{1}{2} + \dfrac{1}{4} - \dfrac{1}{8} + .......$ to 9 terms

Answer

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

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

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

How many terms of the geometric progression 1 + 4 + 16 + 64 + ........ must be added to get sum equal to 5461 ?

Answer

Let n terms be added.

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

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

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

The first term of a G.P. is 27 and its 8^th term is $\dfrac{1}{81}$. Find the sum of its first 10 terms.

Answer

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

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

Since, r < 1

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

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

A boy spends ₹ 10 on first day, ₹ 20 on second day, ₹ 40 on third day and so on. Find how much in all, will he spend in 12 days?

Answer

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

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

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

Hence, ₹10(2¹² - 1).

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

Answer

Given,

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

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

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

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

Dividing (ii) by (i) we get,

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

Substituting value of r in (i) we get,

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

Since, r < 1

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

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

A geometric progression has common ratio = 3 and last term = 486. If the sum of its terms is 728; find its first term.

Answer

Let n^th term be the last term.

⇒ ar^{n - 1} = 486

⇒ a(3)^{n - 1} = 486

⇒ a.3ⁿ.3^-1 = 486

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

Since, r > 1

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

Hence, first term = 2.

Find the sum of G.P. : 3, 6, 12, ......., 1536.

Answer

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

Let 1536 be n^th term

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

Since, r > 1

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

Hence, sum = 3069.

How many terms of the series 2 + 6 + 18 + ...... must be taken to make the sum equal to 728?

Answer

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

Let n be no. of terms taken.

Since, r > 1

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

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

In a G.P., the ratio between the sum of first three terms and that of the first six terms is 125 : 152. Find its common ratio.

Answer

Given,

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

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

If the sum of 1 + 2 + 2² + ........ + 2^{n - 1} is 255, find the value of n.

Answer

Common ratio = 2.

Since, |r| > 1

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

Hence, n = 8.

The sum of three numbers in G.P. is $\dfrac{39}{10}$ and their product is 1. Find the numbers.

Answer

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

Given,

Product = 1.

$\therefore \dfrac{a}{r} \times a \times ar = 1 \\[1em] \Rightarrow a^3 = 1 \\[1em] \Rightarrow a = 1.$

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

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

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

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

a = 1

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

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

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

a = 1

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

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

The first term of a G.P. is -3 and the square of the second term is equal to its 4^th term. Find its 7^th term.

Answer

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.

Find the 5^th term of the G.P. $\dfrac{5}{2}, 1, .......$

Answer

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

a₅ = ar⁴

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

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

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

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

The first two terms of a G.P. are 125 and 25 respectively. Find the 5^th and the 6^th terms of the G.P.

Answer

Given,

⇒ a = 125

⇒ a₂ = 25

⇒ ar = 25

⇒ 125r = 25

⇒ r = $\dfrac{25}{125} = \dfrac{1}{5}$.

a₅ = ar⁴

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

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

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

a₆ = ar⁵

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

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

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

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

Find the sum of the sequence $-\dfrac{1}{3}, 1, -3, 9, .....$ upto 8 terms.

Answer

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

Since, |r| > 1

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

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

The first term of a G.P. is 27. If the 8^th term be $\dfrac{1}{81}$, what will be the sum of 10 terms?

Answer

Given,

⇒ a = 27

⇒ a₈ = ar⁷ = $\dfrac{1}{81}$

⇒ 27r⁷ = $\dfrac{1}{81}$

⇒ r⁷ = $\dfrac{1}{81 \times 27}$

⇒ r⁷ = $\dfrac{1}{3^7}$

⇒ r⁷ = $\Big(\dfrac{1}{3}\Big)^7$

⇒ r = $\dfrac{1}{3}$.

Since, r < 1

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

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

Find a G.P. for which the sum of first two terms is -4 and the fifth term is 4 times the third term.

Answer

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

G.P. = a, ar, ar², ........

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

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. = $-\dfrac{4}{3}, -\dfrac{8}{3}, -\dfrac{16}{3}$........ or 4, -8, 16, -32.