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Class 11 (FYJC / HSC)Mathematics & Statistics2026-27 Syllabus

Chapter 2 Sequences and Series Ex 2.1 Solutions

Complete Maharashtra State Board Balbharati & Yuvakbharati textbook solutions for Chapter 2 Sequences and Series Ex 2.1. Step-by-step solved exercises, numerical problems, and digest answers.

15 Solved Questions23 Diagrams991 words

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

Question 1 Maharashtra Board Solution
Check whether the following sequences are G.P. If so, write tn. (i) 2, 6, 18, 54, ……
Solution & Step-by-Step Answer:

(ii) 1, -5, 25, -125, ………
Solution:

(iii)
Solution:

(iv) 3, 4, 5, 6, ……
Solution:

(v) 7, 14, 21, 28, ……
Solution:

Question 2 Maharashtra Board Solution
For the G.P. (i) If r = , a = 9, find t7.
Solution & Step-by-Step Answer:

(ii) If a = , r = 3, find t6.
Solution:

(iii) If r = -3 and t6= 1701, find a.
Solution:

(iv) If a = , t6= 162, find r.
Solution:

Question 3 Maharashtra Board Solution
Which term of the G. P. 5, 25, 125, 625, …… is 510?
Solution & Step-by-Step Answer:

Question 4 Maharashtra Board Solution
For what values of x, the terms , x, are in G. P.?
Solution & Step-by-Step Answer:

Question 5 Maharashtra Board Solution
If for a sequence, , show that the sequence is a G. P. Find its first term and the common ratio.
Solution & Step-by-Step Answer:

Question 6 Maharashtra Board Solution
Find three numbers in G. P. such that their sum is 21 and the sum of their squares is 189.
Solution & Step-by-Step Answer:
Let the three numbers in G. P. be , a, ar. According to the given conditions, When a = 6, r = 2, = 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 = 122+ 62+ 32
= 144 + 36 + 9
= 189
Thus, our answer is correct.

Question 7 Maharashtra Board Solution
Find four numbers in G. P. such that the sum of the middle two numbers is and their product is 1.
Solution & Step-by-Step Answer:
Let the four numbers in G.P. be According to the given conditions,

Question 8 Maharashtra Board Solution
Find five numbers in G. P. such that their product is 1024 and the fifth term is square of the third term.
Solution & Step-by-Step Answer:
Let the five numbers in G. P. be 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 Maharashtra Board Solution
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 & Step-by-Step Answer:

Question 10 Maharashtra Board Solution
If p, q, r, s are in G.P., show that p + q, q + r, r + s are also in G. P.
Solution & Step-by-Step Answer:
p, q, r, s are in G.P. ∴ Let = 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 ∴ p + q, q + r, r + s are in G.P.

Question 11 Maharashtra Board Solution
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 & Step-by-Step Answer:
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 = = 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 Maharashtra Board Solution
A ball is dropped from a height of 80 ft. The ball is such that it rebounds 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 & Step-by-Step Answer:
Since the ball rebounds of the height it has fallen, the height in successive bounce is in G.P. 1st height in the bounce = 80 ×

Question 13 Maharashtra Board Solution
The numbers 3, x and x + 6 are in G. P. Find (i) x (ii) 20th term (iii) nth term.
Solution & Step-by-Step Answer:
(i) 3, x and x + 6 are in G. P. x2 = 3x + 18 x2 – 3x – 18 = 0 (x – 6) (x + 3) = 0 x = 6, -3

Question 14 Maharashtra Board Solution
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 & Step-by-Step Answer:
a = 200, r = 1 + = Mosquitoes at the end of 1st year = 200 × (i) Number of mosquitoes after 3 years = 200 × = 200 = 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 Maharashtra Board Solution
The numbers x – 6, 2x and x2 are in G. P. Find (i) x (ii) 1st term (iii) nth term
Solution & Step-by-Step Answer:
(i) x – 6, 2x and x are in Geometric progression. ∴ 4x2 = x2(x – 6) 4 = x – 6 x = 10

(ii) t1= x – 6 = 10 – 6 = 4