Maharashtra Board Solutions Class 11-Arts & Science Maths (Part 2): Chapter 2- Sequences and Series

Class 11: Maths Chapter 2 solutions. Complete Class 11 Maths Chapter 2 Notes.

Maharashtra Board Solutions Class 11-Arts & Science Maths (Part 2): Chapter 2- Sequences and Series

Maharashtra Board 11th Maths Chapter 2, Class 11 Maths Chapter 2 solutions

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) Number of mosquitoes after n years = 200 (1.1)ⁿ

Ex 2.2

Question 7.
Find the sum to n terms of the sequence
(i) 0.5, 0.05, 0.005, …..
Solution:

(ii) 0.2, 0.02, 0.002, ……
Solution:

Question 8.
For a sequence, if S_n = 2(3n – 1), find the nth term, hence showing that the sequence is a G.P.
Solution:

Question 9.
If S, P, R are the sum, product, and sum of the reciprocals of n terms of a G.P, respectively, then verify that [SR]n = P².
Solution:
Let a be the 1st term and r be the common ratio of the G.P.
∴ the G.P. is a, ar, ar², ar³, …, ar^n-1

Question 10.
If S_n, S_2n, S_3n are the sum of n, 2n, 3n terms of a G.P. respectively, then verify that S_n(S_3n – S_2n) = (S_2n – S_n)².
Solution:
Let a and r be the 1st term and common ratio of the G.P. respectively.

Ex 2.3

Maharashtra Board Solutions Class 11-Arts & Science Maths (Part 2): Chapter 2- Sequences and Series Ex. 2.3

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

Ex 2.4

Ex 2.5

Solution:

Ex 2.6

Question 8.
Find the sum 1 × 3 × 5 + 3 × 5 × 7 + 5 × 7 × 9 + …… + (2n – 1) (2n + 1) (2n + 3)
Solution:
Let S = 1 × 3 × 5 + 3 × 5 × 7 + ….. upto n terms
Here, 1, 3, 5, 7 … are in A.P. with rth term = 2r – 1,
3, 5, 7, 9,… are in A.P. with rth term = 2r + 1,
5, 7, 9, 11,… are in A.P. with rth term = 2r + 3

Download PDF

Maharashtra Board Solutions Class 11-Arts & Science Maths (Part 2): Chapter 2- Sequences and Series

Download PDF: Maharashtra Board Solutions Class 11-Arts & Science Maths (Part 2): Chapter 2- Sequences and Series PDF

Chapterwise Maharashtra Board Solutions Class 11 Arts & Science Maths (Part 2) :

• Chapter 1- Complex Numbers
• Chapter 2- Sequences and Series
• Chapter 3- Permutations and Combination
• Chapter 4- Methods of Induction and Binomial Theorem
• Chapter 5- Sets and Relations
• Chapter 6- Functions
• Chapter 7- Limits
• Chapter 8- Continuity
• Chapter 9- Differentiation

Read More

• RD Sharma Solutions for Class 11 Maths Chapter 20–Geometric Progressions
• RD Sharma Solutions for Class 11 Maths Chapter 21–Some Special Series
• RD Sharma Solutions for Class 7 Maths: Chapter 6–Exponents
• Maharashtra Board Solutions Class 11-Arts & Science Maths (Part 1): Chapter 4- Determinants and Matrices
• Maharashtra Board Solutions Class 11-Arts & Science Maths (Part 1): Chapter 3- Trigonometry – II