Question

Q.17 \( \quad \) Sum \( \frac{1}{5}+\frac{1}{7}+\frac{1}{5^{2}}+\frac{1}{7^{2}}+\ldots . \) to \( \infty= \) (A) \( 5 / 12 \) (B) \( 3 / 4 \) (C) \( 7 / 12 \) (D) \( 3 / 49 \)

Answer 1

1/5 + 1/7 + 1/5² + 1/7² + ....

= (1/5 + 1/5² + 1/5³ + ...) + (1/7 + 1/7² + 1/7³ + ...)

\(=\cfrac{\frac{1}{5}}{1-\frac{1}{5}}+\cfrac{\frac{1}{7}}{1-\frac{1}{7}}\) (∵ Sum of infinite series in G.P. when r < 1 is a/1-r)

\(=\cfrac{\frac{1}{5}}{\frac{4}{5}}+\cfrac{\frac{1}{7}}{\frac{6}{7}}\)

= 1/4 + 1/6 \(=\frac{3+2}{12}\) = 5/12.