Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 2 Sequences and Series Ex 2.3 Questions and Answers.

## Maharashtra State Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.3

Question 1.

Determine whether the sum to infinity of the following G.P.s exist, if exists find them.

(i) \(\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \ldots\)

Solution:

(ii) \(2, \frac{4}{3}, \frac{8}{9}, \frac{16}{27}, \ldots\)

Solution:

(iii) \(-3,1, \frac{-1}{3}, \frac{1}{9}, \ldots\)

Solution:

(iv) \(\frac{1}{5}, \frac{-2}{5}, \frac{4}{5}, \frac{-8}{5}, \frac{16}{5}, \ldots\)

Solution:

(v) 9, 8.1, 7.29, ……

Solution:

9, 8.1, 7.29, …..

Here, a = 9, r = \(\frac{8.1}{9}\) = 0.9, |r| < 1

∴ Sum to infinity exists.

∴ Sum to infinity = \(\frac{\mathrm{a}}{1-\mathrm{r}}\)

= \(\frac{9}{1-0.9}\)

= \(\frac{9}{0.1}\)

= 90

Question 2.

Express the following recurring decimals as rational numbers.

(i) \(0 . \overline{7}\)

(ii) \(2 . \overline{4}\)

(iii) \(2.3 \overline{5}\)

(iv) \(51.0 \overline{2}\)

Solution:

(i) \(0 . \overline{7}\) = 0.7777… = 0.7 + 0.07 + 0.007 + ….

The terms 0.7, 0.07, 0.007,… are in G.P.

∴ a = 0.7, r = \(\frac{0.07}{0.7}\) = 0.1, |r| = |0.1| < 1

∴ Sum to infinity exists.

∴ Sum to infinity

(ii) \(2 . \overline{4}\) = 2.444 … = 2 + 0.4 + 0.04 + 0.004 + …

The terms 0.4, 0.04, 0.004,… are in G.P.

∴ a = 0.4, r = \(\frac{0.07}{0.7}\) = 0.1, |r| = 10.11 < 1

∴ Sum to infinity exists.

∴ Sum to infinity

(iii) \(2.3 \overline{5}\) = 2.3555… = 2.3 + 0.05 + 0.005 + 0.0005 + …

The terms 0.05,0.005,0.0005,… are in G.P.

∴ a = 0.05, r = \(\frac{0.005}{0.05}\) = 0.1, |r| = |0.1| < 1

∴ Sum to infinity exists.

∴ Sum to infinity

(iv) \(51.0 \overline{2}\) = 51.0222 …. = 51 + 0.02 + 0.002 + 0.0002 + …..

The terms 0.02, 0.002, 0.0002,… are in G.P.

∴ a = 0.02, r = \(\frac{0.002}{0.02}\) = 0.1, |r| = |0.1| < 1

∴ Sum to infinity exists.

∴ Sum to infinity

Question 3.

If the common ratio of a G.P. is \(\frac{2}{3}\) and the sum to infinity is 12, find the first term.

Solution:

r = \(\frac{2}{3}\), sum to infinity = 12 ….. [Given]

Sum to infinity = \(\frac{\mathrm{a}}{1-\mathrm{r}}\)

12 = \(\frac{a}{1-\frac{2}{3}}\)

a = 12 × \(\frac{1}{3}\)

∴ a = 4

Question 4.

If the first term of the G.P. is 6 and its sum to infinity is \(\frac{96}{17}\), find the common ratio.

Solution:

a = 6, sum to infinity = \(\frac{96}{17}\) …..[Given]

Sum to infinity = \(\frac{\mathrm{a}}{1-\mathrm{r}}\)

Question 5.

The sum of an infinite G.P. is 5 and the sum of the squares of these terms is 15, find the G.P.

Solution:

Let the required G.P. be a, ar, ar^{2}, ar^{3}, …..

Sum to infinity of this G.P. = 5

Question 6.

Find

(i) \(\sum_{r=1}^{\infty} 4(0.5)^{r}\)

Solution:

(ii) \(\sum_{r=1}^{\infty}\left(-\frac{1}{3}\right)^{r}\)

Solution:

(iii) \(\sum_{r=0}^{\infty}(-8)\left(-\frac{1}{2}\right)^{r}\)

Solution:

(iv) \(\sum_{n=1}^{\infty} 0.4^{n}\)

Solution:

Question 7.

The midpoints of the sides of a square of side 1 are joined to form a new square. This procedure is repeated indefinitely. Find the sum of

(i) the areas of all the squares.

(ii) the perimeters of all the squares.

Solution:

(i) Area of the 1st square = 12

Area of the 2nd square = \(\left(\frac{1}{\sqrt{2}}\right)^{2}\)

Area of the 3rd square = \(\left(\frac{1}{2}\right)^{2}\)

and so on.

∴ Sum of the areas of all the squares

∴ Sum to infinity exists.

∴ Sum of the areas of all the squares = \(\frac{1}{1-\frac{1}{2}}\) = 2

(ii) Perimeter of 1st square = 4

Perimeter of 2nd square = 4\(\left(\frac{1}{\sqrt{2}}\right)\)

Perimeter of 3rd square = 4\(\left(\frac{1}{2}\right)\)

and so on.

Sum of the perimeters of all the squares

Question 8.

A ball is dropped from a height of 10 m. It bounces to a height of 6m, then 3.6 m, and so on. Find the total distance travelled by the ball.

Solution:

Here, on the first bounce, the ball will go 6 m and it will return 6 m.

On the second bounce, the ball will go 3.6 m and it will return 3.6 m, and so on.

Given that, a ball is dropped from a height of 10 m.

∴ Total distance travelled by the ball is = 10 + 2[6 + 3.6 + …]

The terms 6, 3.6 … are in G.P.

a = 6, r = 0.6, |r| = |0.6| < 1

∴ Sum to infinity exists.

∴ Total distance travelled by the ball