Chapter 2 Sequences and Series Ex 2.2 Solutions

Maharashtra State Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.2

(ii) p, q,
Solution:

(iii) 0.7, 0.07, 0.007, …….
Solution:

(iv) √5, -5, 5√5, -25, …….
Solution:

(ii) If S₅= 1023, r = 4, find a.
Solution:

(ii) For a G.P. sum of the first 3 terms is 125 and the sum of the next 3 terms is 27, find the value of r.
Solution:

(ii) If t₄= 16, t₉= 512, find S₁₀.
Solution:

(ii) 8 + 88 + 888 + 8888 + …..
Solution:
S_n= 8 + 88 + 888 + … upto n terms
= 8(1 + 11 + 111 + … upto n terms)
= (9 + 99 + 999 + … upto n terms)
= [(10 – 1) + (100 – 1) + (1000 – 1) +… upto n terms]
= [(10 + 100 + 1000 + … upto n terms) – (1 + 1 + 1 + … n times)]
But 10, 100, 1000, … n terms are in G.P. with
a = 10, r = = 10

(ii) 0.7 + 0.77 + 0.777 + ……
Solution:
S_n= 0.7 + 0.77 + 0.777 + … upto n terms
= 7(0.1 + 0.11 + 0.111 + … upton terms)
= (0.9 + 0.99 + 0.999 + … upto n terms)
= [(1 – 0.1) + (1 – 0.01) + (1 – 0.001) +… upto n terms]
= [(1 + 1 + 1 +… n times) – (0.1 + 0.01 + 0.001 +… upto n terms )]
But 0.1, 0.01, 0.001, … n terms are in G.P. with
a = 0.1, r = = 0.1

(ii) 0.2, 0.02, 0.002, ……
Solution:

(ii)
Solution: