Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 2 Sequences and Series Ex 2.1 Questions and Answers.

## Maharashtra State Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.1

Question 1.

Check whether the following sequences are G.P. If so, write t_{n}.

(i) 2, 6, 18, 54, ……

Solution:

(ii) 1, -5, 25, -125, ………

Solution:

(iii) \(\sqrt{5}, \frac{1}{\sqrt{5}}, \frac{1}{5 \sqrt{5}}, \frac{1}{25 \sqrt{5}}, \cdots\)

Solution:

(iv) 3, 4, 5, 6, ……

Solution:

(v) 7, 14, 21, 28, ……

Solution:

Question 2.

For the G.P.

(i) If r = \(\frac{1}{3}\), a = 9, find t_{7}.

Solution:

(ii) If a = \(\frac{7}{243}\), r = 3, find t_{6}.

Solution:

(iii) If r = -3 and t_{6} = 1701, find a.

Solution:

(iv) If a = \(\frac{2}{3}\), t_{6} = 162, find r.

Solution:

Question 3.

Which term of the G. P. 5, 25, 125, 625, …… is 5^{10}?

Solution:

Question 4.

For what values of x, the terms \(\frac{4}{3}\), x, \(\frac{4}{27}\) are in G. P.?

Solution:

Question 5.

If for a sequence, \(\mathrm{t}_{\mathrm{n}}=\frac{5^{n-3}}{2^{n-3}}\), show that the sequence is a G. P. Find its first term and the common ratio.

Solution:

Question 6.

Find three numbers in G. P. such that their sum is 21 and the sum of their squares is 189.

Solution:

Let the three numbers in G. P. be \(\frac{a}{r}\), a, ar.

According to the given conditions,

When a = 6, r = 2,

\(\frac{a}{r}\) = 3, a = 6, ar = 12

Hence, the three numbers in G.P. are 12, 6, 3 or 3, 6, 12.

Check:

If sum of the three numbers is 21 and sum of their squares is 189, then our answer is correct.

Sum of the numbers = 12 + 6 + 3 = 21

Sum of the squares of the numbers = 12^{2} + 6^{2} + 3^{2}

= 144 + 36 + 9

= 189

Thus, our answer is correct.

Question 7.

Find four numbers in G. P. such that the sum of the middle two numbers is \(\frac{10}{3}\) and their product is 1.

Solution:

Let the four numbers in G.P. be \(\frac{a}{r^{3}}, \frac{a}{r}, a r, a r^{3}\)

According to the given conditions,

Question 8.

Find five numbers in G. P. such that their product is 1024 and the fifth term is square of the third term.

Solution:

Let the five numbers in G. P. be

\(\frac{a}{r^{2}}, \frac{a}{r}, a, a r, a r^{2}\)

According to the given conditions,

Hence, the five numbers in G.P. are

1, 2, 4, 8, 16 or 1, -2, 4, -8, 16.

Question 9.

The fifth term of a G. P. is x, the eighth term of a G.P. is y and the eleventh term of a G.P. is z, verify whether y^{2} = xz.

Solution:

Question 10.

If p, q, r, s are in G.P., show that p + q, q + r, r + s are also in G. P.

Solution:

p, q, r, s are in G.P.

∴ \(\frac{\mathrm{q}}{\mathrm{p}}=\frac{\mathrm{r}}{\mathrm{q}}=\frac{\mathrm{s}}{\mathrm{r}}\)

Let \(\frac{\mathrm{q}}{\mathrm{p}}=\frac{\mathrm{r}}{\mathrm{q}}=\frac{\mathrm{s}}{\mathrm{r}}\) = k

∴ q = pk, r = qk, s = rk

We have to prove that p + q, q + r, r + s are in G.P.

i.e., to prove that \(\frac{\mathrm{q}+\mathrm{r}}{\mathrm{p}+\mathrm{q}}=\frac{\mathrm{r}+\mathrm{s}}{\mathrm{q}+\mathrm{r}}\)

∴ p + q, q + r, r + s are in G.P.

Question 11.

The number of bacteria in a culture doubles every hour. If there were 50 bacteria originally in the culture, how many bacteria will be there at the end of the 5th hour?

Solution:

Since the number of bacteria in culture doubles every hour, increase in number of bacteria after every hour is in G.P.

∴ a = 50, r = \(\frac{100}{50}\) = 2

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

To find the number of bacteria at the end of the 5th hour.

(i.e., to find the number of bacteria at the beginning of the 6th hour, i.e., to find t_{6}.)

∴ t_{6} = ar^{5}

= 50 × (2^{5})

= 50 × 32

= 1600

Question 12.

A ball is dropped from a height of 80 ft. The ball is such that it rebounds \(\left(\frac{3}{4}\right)^{\text {th }}\) of the height it has fallen. How high does the ball rebound on the 6th bounce? How high does the ball rebound on the nth bounce?

Solution:

Since the ball rebounds \(\left(\frac{3}{4}\right)^{\text {th }}\) of the height it has fallen, the height in successive bounce is in G.P.

1st height in the bounce = 80 × \(\frac{3}{4}\)

Question 13.

The numbers 3, x and x + 6 are in G. P. Find

(i) x

(ii) 20th term

(iii) nth term.

Solution:

(i) 3, x and x + 6 are in G. P.

\(\frac{x}{3}=\frac{x+6}{x}\)

x^{2} = 3x + 18

x^{2} – 3x – 18 = 0

(x – 6) (x + 3) = 0

x = 6, -3

Question 14.

Mosquitoes are growing at a rate of 10% a year. If there were 200 mosquitoes in the beginning, write down the number of mosquitoes after

(i) 3 years

(ii) 10 years

(iii) n years

Solution:

a = 200, r = 1 + \(\frac{10}{100}\) = \(\frac{11}{10}\)

Mosquitoes at the end of 1st year = 200 × \(\frac{11}{10}\)

(i) Number of mosquitoes after 3 years

= 200 × \(\frac{11}{10} \times\left(\frac{11}{10}\right)^{2}\)

= 200 \(\left(\frac{11}{10}\right)^{3}\)

= 200 (1.1)^{3}

(ii) Number of mosquitoes after 10 years = 200 (1.1)^{10}

(iii) Number of mosquitoes after n years = 200 (1.1)^{n}

Question 15.

The numbers x – 6, 2x and x^{2} are in G. P. Find

(i) x

(ii) 1st term

(iii) nth term

Solution:

(i) x – 6, 2x and x are in Geometric progression.

∴ \(\frac{2 x}{x-6}=\frac{x^{2}}{2 x}\)

4x^{2} = x^{2}(x – 6)

4 = x – 6

x = 10

(ii) t_{1} = x – 6 = 10 – 6 = 4