Geometric Progression — Exercise 11(B)

Find the sum of first 8 terms of the G.P. 1, 3, 9, 27, 81, .....

Answer

Given,

a = 1

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

n = 8

We know that,

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

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

Substituting values we get :

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

Hence, S₈ = 3280.

Find the sum of first 10 terms of the G.P. 1, $\sqrt{3}$, 3, $3\sqrt{3}$, ………

Answer

Given,

a = 1

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

n = 10

We know that,

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

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

Substituting values we get :

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

Rationalizing the Denominator :

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

Hence, S₁₀ = $121(\sqrt3 + 1)$.

Find the sum of first 9 terms of the G.P. 1, $-\dfrac{1}{2}$, $\dfrac{1}{4}$, $-\dfrac{1}{8}$, ………

Answer

Given,

a = 1

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

n = 9

We know that,

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

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

Substituting values we get :

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

Hence, S₉ = $\dfrac{171}{256}$.

Find the sum of first 6 terms of the G.P. 0.1, 0.01, 0.001, .....

Answer

Given,

a = 0.1

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

n = 6

We know that,

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

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

Substituting values we get :

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

Hence, S₆ = 0.111111.

Find the sum of first 6 terms of the G.P. 1, $-\dfrac{1}{3}, \dfrac{1}{3^{2}}, -\dfrac{1}{3^{3}}$, .......

Answer

Given,

a = 1

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

n = 6

We know that,

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

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

Substituting values we get :

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

Hence, S₆ = $\dfrac{182}{243}$.

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

Answer

Given,

a = 27

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

We know that,

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

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

Substituting values we get :

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

Hence, S₇ = $\dfrac{1093}{27}$.

The 4th and the 7th terms of a G.P. are $\dfrac{1}{27}$ and $\dfrac{1}{729}$ respectively. Find the sum of first 6 terms of the G.P.

Answer

Given,

⇒ T₄ = $\dfrac{1}{27}$

⇒ ar³ = $\dfrac{1}{27}$ .....(1)

Given,

⇒ T₇ = $\dfrac{1}{729}$

⇒ ar⁶ = $\dfrac{1}{729}$ ..........(2)

Dividing Equation (2) by Equation (1) :

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

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

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

We know that,

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

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

Substituting values we get :

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

Hence, S₆ = $\dfrac{364}{243}$.

If the 6th term of a series in Geometric Progression (G.P.) is 32 and the 9th term is 256, find the:

(i) first term and the common ratio.

(ii) sum of its first 10 terms.

Answer

(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:

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

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

$\Rightarrow a \times 2^5 = 32 \\[1em] \Rightarrow a \times 32 = 32 \\[1em] \Rightarrow a = \dfrac{32}{32} \\[1em] \Rightarrow a = 1.$

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

(ii) By formula,

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

Substituting values we get :

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

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

15, 30, 60, 120 ...... are in G.P. (Geometric Progression).

(i) Find the nth term of this G.P. in terms of n.

(ii) How many terms of the above G.P. will give the sum 945 ?

Answer

Given,

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

First term (a) = 15

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

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

= $\dfrac{15}{2}$ 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. = $\dfrac{a(r^n - 1)}{(r - 1)}$

Substituting values we get :

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

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

How many terms of the G.P. $\dfrac{2}{9}$, $-\dfrac{1}{3}$, $\dfrac{1}{2}$, ……… must be taken to make the sum equal to $\dfrac{55}{72}$?

Answer

In the given G.P.,

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

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

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

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

We know that,

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

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

Substituting values we get :

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

Hence, n = 5.

In a G.P. the ratio of the sum of first 3 terms is to that of first 6 terms is 125 : 152. Find the common ratio.

Answer

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

Given,

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

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

A manufacturer reckons that the value of a machine which costs him ₹ 31,250 depreciates each year by 20%. Find its value after 2 years.

Answer

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

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

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

Find the sum of the following to n terms: 7 + 77 + 777 + 7777 + ………

Answer

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

Now,

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

a = 10

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

We know that,

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

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

$\Rightarrow S_n = \dfrac{10(10^n - 1)}{10 - 1} \\[1em] = \dfrac{10(10^n - 1)}{9}.$

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

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

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

Find the sum of n terms of series whose mth term is 2^m + 2m

Answer

Given,

mth term = 2^m + 2m

1st term = 2¹ + 2 × 1

2nd term = 2² + 2 × 2

nth term = 2ⁿ + 2 × n

$S_n = [(2^1 + 2 \times 1) + (2^2 + 2 \times 2) +(2^3 + 2 \times 3)+........ + (2^n + 2 \times n)] \\[1em] = [2^1 + 2^2 + 2^3 + ..... 2^n + (2 \times 1 + 2 \times 2 + 2 \times 3 + ..... + 2 \times n)] \\[1em] = (2^1 + 2^2 + 2^3 + ..... + 2^n) + 2(1 + 2 + 3 + ..... + n)......(1)$

Calculating :

2¹ + 2² + ........ + 2ⁿ

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

By using formula,

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

Substitute values, we get:

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

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

Substitute values, we get:

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

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

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

Substitute values in (1) we get,

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

= 2(2ⁿ - 1) + n(n + 1)

Hence, S_n = 2(2ⁿ - 1) + n(n + 1).