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 tn.
(i) 2, 6, 18, 54, ……
Solution:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.1 Q1 (i)

(ii) 1, -5, 25, -125, ………
Solution:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.1 Q1 (ii)

(iii) \(\sqrt{5}, \frac{1}{\sqrt{5}}, \frac{1}{5 \sqrt{5}}, \frac{1}{25 \sqrt{5}}, \cdots\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.1 Q1 (iii)

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

(iv) 3, 4, 5, 6, ……
Solution:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.1 Q1 (iv)

(v) 7, 14, 21, 28, ……
Solution:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.1 Q1 (v)

Question 2.
For the G.P.
(i) If r = \(\frac{1}{3}\), a = 9, find t7.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.1 Q2 (i)

(ii) If a = \(\frac{7}{243}\), r = 3, find t6.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.1 Q2 (ii)

(iii) If r = -3 and t6 = 1701, find a.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.1 Q2 (iii)

(iv) If a = \(\frac{2}{3}\), t6 = 162, find r.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.1 Q2 (iv)

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

Question 3.
Which term of the G. P. 5, 25, 125, 625, …… is 510?
Solution:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.1 Q3

Question 4.
For what values of x, the terms \(\frac{4}{3}\), x, \(\frac{4}{27}\) are in G. P.?
Solution:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.1 Q4

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:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.1 Q5
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.1 Q5.1

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,
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.1 Q6
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.1 Q6.1
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.

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

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 = 122 + 62 + 32
= 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,
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.1 Q7
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.1 Q7.1

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,
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.1 Q8
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 y2 = xz.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.1 Q9

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 \(\frac{\mathrm{q}+\mathrm{r}}{\mathrm{p}+\mathrm{q}}=\frac{\mathrm{r}+\mathrm{s}}{\mathrm{q}+\mathrm{r}}\)
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.1 Q10
∴ p + q, q + r, r + s are in G.P.

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

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
tn = arn-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 t6.)
∴ t6 = ar5
= 50 × (25)
= 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 × \(\frac{3}{4}\)
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.1 Q12

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.
\(\frac{x}{3}=\frac{x+6}{x}\)
x2 = 3x + 18
x2 – 3x – 18 = 0
(x – 6) (x + 3) = 0
x = 6, -3
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.1 Q13

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

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 x2 are in G. P. Find
(i) x
(ii) 1st term
(iii) nth term
Solution:
(i) x – 6, 2x and x are in Geometric progression.
∴ \(\frac{2 x}{x-6}=\frac{x^{2}}{2 x}\)
4x2 = x2(x – 6)
4 = x – 6
x = 10

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

(ii) t1 = x – 6 = 10 – 6 = 4

Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.1 Q15

Leave a Reply