Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 4 Sequences and Series Ex 4.2 Questions and Answers.

## Maharashtra State Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.2

Question 1.

For the following G.P.’s, find S_{n}.

(i) 3, 6, 12, 24, …..

(ii) \(\mathbf{p}, \mathbf{q}, \frac{\mathbf{q}^{2}}{\mathbf{p}}, \frac{\mathbf{q}^{3}}{\mathbf{p}^{2}}, \ldots\)

Solution:

Question 2.

For a G.P., if

(i) a = 2, r = \(-\frac{2}{3}\), find S_{6}.

(ii) S_{5} = 1023, r = 4, find a.

Solution:

Question 3.

For a G. P., if

(i) a = 2, r = 3, S_{n} = 242, find n.

(ii) sum of the first 3 terms is 125 and the sum of the next 3 terms is 27, find the value of r.

Solution:

(i) a = 2, r = 3, S_{n} = 242

Sn = \(a\left(\frac{r^{n}-1}{r-1}\right)\), for r > 1

Question 4.

For a G. P.,

(i) if t_{3} = 20, t_{6} = 160, find S_{7}.

(ii) if t_{4} = 16, t_{9} = 512, find S_{10}.

Solution:

(i) t_{3} = 20, t_{6} = 160

t_{n} = ar^{n-1}

∴ t_{3} = ar^{3-1} = ar^{2}

∴ ar^{2} = 20

∴ a = \(\frac{20}{\mathrm{r}^{2}}\) ……(i)

Question 5.

Find the sum to n terms:

(i) 3 + 33 + 333 + 3333 + ……

(ii) 8 + 88 + 888 + 8888 + ……..

Solution:

(i) S_{n} = 3 + 33 + 333 +….. upto n terms

= 3(1 + 11 + 111 +….. upto n terms)

= \(\frac{3}{9}\)(9 + 99 + 999 + … upto n terms)

= \(\frac{3}{9}\)[(10 – 1) + (100 – 1) + (1000 – 1) +… upto n terms]

= \(\frac{3}{9}\)[(10 + 100 + 1000 + … upto n terms) – (1 + 1 + 1 + … n times)]

But 10, 100, 1000, … n terms are in G.P.

with a = 10, r = \(\frac{100}{10}\) = 10

(ii) S_{n} = 8 + 88 + 888 + … upto n terms

= 8(1 + 11 + 111 + … upto n terms)

= \(\frac{8}{9}\) (9 + 99 + 999 + … upto n terms)

= \(\frac{8}{9}\) [(10 – 1) + (100 – 1) + (1000 – 1) +… upto n terms]

= \(\frac{8}{9}\) [(10 + 100 + 1000 + … upto n terms) – (1 + 1 + 1 + … n times)]

But 10, 100, 1000, … n terms are in G.P. with

a = 10, r = \(\frac{100}{10}\) = 10

Question 6.

Find the sum to n terms:

(i) 0.4 + 0.44 + 0.444 + ……

(ii) 0.7 + 0.77 + 0.777 + …..

Solution:

(i) S_{n} = 0.4 + 0.44 + 0.444 + ….. upto n terms

= 4(0.1 + 0.11 + 0.111 + …. upto n terms)

= \(\frac{4}{9}\) (0.9 + 0.99 + 0.999 + … upto n terms)

= \(\frac{4}{9}\) [(i – 0.1) + (1 – 0.01) + (1 – 0.001) … upto n terms]

= \(\frac{4}{9}\) [(1 + 1 + 1 + …n times) – (0.1 + 0.01 + 0.001 +… upto n terms)]

But 0.1, 0.01, 0.001, … n terms are in G.P.

with a = 0.1, r = \(\frac{0.01}{0.1}\) = 0.1

∴ S_{n} = \(\frac{4}{9}\left\{\mathrm{n}-0.1\left[\frac{1-(0.1)^{\mathrm{n}}}{1-0.1}\right]\right\}\)

(ii) S_{n} = 0.7 + 0.77 + 0.777 + … upto n terms

= 7(0.1 + 0.11 + 0.111 + … upto n terms)

= \(\frac{7}{9}\) (0.9 + 0.99 + 0.999 + … upto n terms)

= \(\frac{7}{9}\) [(1 – 0.1) + (1 – 0.01) + (1 – 0.001) +… upto n terms]

= \(\frac{7}{9}\) [(1 + 1 + 1 +… n times) – (0.1 + 0.01 + 0.001 +… upto n terms)]

But 0.1, 0.01, 0.001, … n terms are in G.P.

with a = 0.1, r = \(\frac{0.01}{0.1}\) = 0.1

Question 7.

Find the nth terms of the sequences:

(i) 0.5, 0.55, 0.555,…..

(ii) 0.2, 0.22, 0.222,…..

Solution:

(i) Let t_{1} = 0.5, t_{2} = 0.55, t_{3} = 0.555 and so on.

t_{1} = 0.5

t_{2} = 0.55 = 0.5 + 0.05

t_{3} = 0.555 = 0.5 + 0.05 + 0.005

∴ t_{n} = 0.5 + 0.05 + 0.005 + … upto n terms

But 0.5, 0.05, 0.005, … upto n terms are in G.P. with a = 0.5 and r = 0.1

∴ t_{n} = the sum of first n terms of the G.P.

(ii) Let t_{1} = 0.2, t_{2} = 0.22, t_{3} = 0.222 and so on

t_{1} = 0.2

t_{2} = 0.22 = 0.2 + 0.02

t_{3} = 0.222 = 0.2 + 0.02 + 0.002

∴ t_{n} = 0.2 + 0.02 + 0.002 + … upto n terms

But 0.2, 0.02, 0.002, … upto n terms are in G.P. with a = 0.2 and r = 0.1

∴ t_{n} = the sum of first n terms of the G.P.

Question 8.

For a sequence, if S_{n} = 2(3^{n-1}), find the nth term, hence showing that the sequence is a G.P.

Solution:

Question 9.

If S, P, R are the sum, product and sum of the reciprocals of n terms of a G.P. respectively, then verify that \(\left(\frac{\mathbf{S}}{\mathbf{R}}\right)^{\mathbf{n}}\) = P^{2}.

Solution:

Let a be the 1st term and r be the common ratio of the G.P.

∴ the G.P. is a, ar, ar^{2}, ar^{3}, …, ar^{n-1}

Question 10.

If S_{n}, S_{2n}, S_{3n} are the sum of n, 2n, 3n terms of a G.P. respectively, then verify that S_{n} (S_{3n} – S_{2n}) = (S_{2n} – S_{n})^{2}.

Solution:

Let a and r be the 1st term and common ratio of the G.P. respectively.