Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 2 Sequences and Series Ex 2.2 Questions and Answers.

## Maharashtra State Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.2

Question 1.

For the following G.P.s, find S_{n}.

(i) 3, 6, 12, 24, ……..

Solution:

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

Solution:

(iii) 0.7, 0.07, 0.007, …….

Solution:

(iv) √5, -5, 5√5, -25, …….

Solution:

Question 2.

For a G.P.

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

Solution:

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

Solution:

Question 3.

For a G.P.

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

Solution:

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

Solution:

Question 4.

For a G.P.

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

Solution:

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

Solution:

Question 5.

Find the sum to n terms

(i) 3 + 33 + 333 + 3333 + …..

Solution:

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 nterms) – (1 + 1 + 1 + ….. n times)]

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

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

(ii) 8 + 88 + 888 + 8888 + …..

Solution:

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 + …..

Solution:

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)

= \(\frac{4}{9}\)[(1 – 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

(ii) 0.7 + 0.77 + 0.777 + ……

Solution:

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

= 7(0.1 + 0.11 + 0.111 + … upton 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 sum to n terms of the sequence

(i) 0.5, 0.05, 0.005, …..

Solution:

(ii) 0.2, 0.02, 0.002, ……

Solution:

Question 8.

For a sequence, if S_{n} = 2(3n – 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 \(\left[\frac{S}{R}\right]^{n}\) = 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.

Question 11.

Find

(i) \(\sum_{r=1}^{10}\left(3 \times 2^{r}\right)\)

Solution:

(ii) \(\sum_{r=1}^{10} 5 \times 3^{r}\)

Solution:

Question 12.

The value of a house appreciates 5% per year. How much is the house worth after 6 years if its current worth is Rs. 15 Lac. [Given: (1.05)^{5} = 1.28, (1.05)^{6} = 1.34]

Solution:

The value of a house is Rs. 15 Lac.

Appreciation rate = 5% = \(\frac{5}{100}\) = 0.05

Value of house after 1st year = 15(1 + 0.05) = 15(1.05)

Value of house after 6 years = 15(1.05) (1.05)^{5}

= 15(1.05)^{6}

= 15(1.34)

= 20.1 lac.

Question 13.

If one invests Rs. 10,000 in a bank at a rate of interest 8% per annum, how long does it take to double the money by compound interest? [(1.08)^{5} = 1.47]

Solution:

Amount invested = Rs. 10000

Interest rate = \(\frac{8}{100}\) = 0.08

amount after 1st year = 10000(1 + 0.08) = 10000(1.08)

Value of the amount after n years

= 10000(1.08) × (1.08)^{n-1}

= 10000(1.08)^{n}

= 20000

∴ (1.08)^{n} = 2

∴ (1.08)^{5} = 1.47 …..[Given]

∴ n = 10 years, (approximately)