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

Chapter 4 Sequences and Series Ex 4.1 Solutions

Complete Maharashtra State Board Balbharati & Yuvakbharati textbook solutions for Chapter 4 Sequences and Series Ex 4.1. Step-by-step solved exercises, numerical problems, and digest answers.

10 Solved Questions14 Diagrams752 words

Balbharati Maharashtra State Board11th Commerce Maths Solution Book PdfChapter 4 Sequences and Series Ex 4.1 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.1

Question 1 Maharashtra Board Solution
Verify whether the following sequences are G.P. If so, write tn. (i) 2, 6, 18, 54, …… (ii) 1, -5, 25, -125, ……. (iii) (iv) 3, 4, 5, 6, …… (v) 7, 14, 21, 28, …..
Solution & Step-by-Step Answer:
(i) 2, 6, 18, 54, ……. t1 = 2, t2 = 6, t3 = 18, t4 = 54, ….. Here, Since, the ratio of any two consecutive terms is a constant, the given sequence is a geometric progression. Here, a = 2, r = 3 tn= arn-1 ∴ tn = 2(3n-1)

(ii) 1, -5, 25, -125, ……
t1= 1, t2= -5, t3= 25, t4= -125, …..
Here,
Since, the ratio of any two consecutive terms is a constant, the given sequence is a geometric progression.
Here, a = 1, r = -5
tn= arn-1
∴ tn= (-5)n-1

(iii)

Since, the ratio of any two consecutive terms is a constant, the given sequence is a geometric progression.

(iv) 3, 4, 5, 6,……
t1= 3, t2= 4, t3= 5, t4= 6, …..
Here,
Since,
∴ the given sequence is not a geometric progression.

(v) 7, 14, 21, 28, …..
t1= 7, t2= 14, t3= 21, t4= 28, …..
Here,
Since,
∴ the given sequence is not a geometric progression.

Question 2 Maharashtra Board Solution
For the G.P., (i) if r = , a = 9, find t7. (ii) if a = , r = , find t3. (iii) if a = 7, r = -3, find t6. (iv) if a = , t6 = 162, find r.
Solution & Step-by-Step Answer:

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, , 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:
The sequence (tn) is a G.P., if = constant, for all n ∈ N ∴ the sequence is a G. P. with common ratio First term, t1 =

Question 6 Maharashtra Board Solution
Find three numbers in G. P. such that their sum is 21 and sum of their squares is 189.
Solution & Step-by-Step Answer:
Let the three numbers in G. P. be , a, ar. According to the first condition, ∴ the three numbers are 12, 6, 3 or 3, 6, 12. Check: First condition: 12, 6, 3 are in G.P. with r = 12 + 6 + 3 = 21 Second condition: 122 + 62 + 32 = 144 + 36 + 9 = 189 Thus, both the conditions are satisfied.

Question 7 Maharashtra Board Solution
Find four numbers in G. P. such that 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 second condition, ∴ a4 = 1 ∴ a = 1 According to the first condition,

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, When a = 4, r = -2 = 1, = -2, a = 4, ar = -8, ar2 = 16 ∴ 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, eighth term of the G. P. is y and eleventh term of the 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. ∴ p + q, q + r, r + s are in G.P.