1 + (0.04) + (0.04)2 + ………………. ∞ = A) 24/25B) 1 C) 0.0

1.	1 + (0.04) + (0.04)2 + ………………. ∞ = A) 24/25B) 1 C) 0.04 D) 25/24
Answer» Correct option is (D) 25/24 \(\because1,0.04,(0.04)^2,.......\) will form a G.P. whose first term is a = 1 & common difference is r = 0.04 < 1 \(\therefore S_\infty=1+0.04+(0.04)^2+……\) \(=\frac1{1-0.04}\) \((\because S_\infty=\frac a{1-r},r<1)\) \(=\frac1{0.96}\) \(=\frac{100}{96}\) \(=\frac{25}{24}\) Hence, \(1+0.04+(0.04)^2+……\) \(=\frac{25}{24}\) Correct option is D) 25/24

1 + (0.04) + (0.04)2 + ………………. ∞ = A) 24/25B) 1 C) 0.04 D) 25/24

Answer»

Correct option is (D) 25/24

\(\because1,0.04,(0.04)^2,.......\) will form a G.P. whose first term is a = 1 & common difference is r = 0.04 < 1

\(\therefore S_\infty=1+0.04+(0.04)^2+……\)

\(=\frac1{1-0.04}\) \((\because S_\infty=\frac a{1-r},r<1)\)

\(=\frac1{0.96}\) \(=\frac{100}{96}\)

\(=\frac{25}{24}\)

Hence, \(1+0.04+(0.04)^2+……\) \(=\frac{25}{24}\)

Correct option is D) 25/24