Question

The sum of the infinite series $1+(1+\frac{1}{5})\left(\frac{1}{2}\right)+(1+\frac{1}{5}+\frac{1}{5^2})\left(\frac{1}{2^2}\right)+.......$

• A. $\frac{20}{9}$
• B. $\frac{10}{9}$
• C. $\frac{5}{9}$
• D. $\frac{5}{3}$

Answer 1

The correct option is A $\frac{20}{9}$

Let $S=1+(1+\frac{1}{5})\left(\frac{1}{2}\right)+(1+\frac{1}{5}+\frac{1}{5^2})\left(\frac{1}{2^2}\right)+...$

$\Rightarrow \frac{S}{2}=\frac{1}{2}+(1+\frac{1}{5})\left(\frac{1}{2^2}\right)+(1+\frac{1}{5}+\frac{1}{5^2})\left(\frac{1}{2^3}\right)+...$

$S-\frac{S}{2}=1+\frac{1}{5}\left(\frac{1}{2}\right)+\frac{1}{5^2}\left(\frac{1}{2^2}\right)+\left(\frac{1}{5^3}\right)\left(\frac{1}{2^3}\right)+...$

$\Rightarrow \frac{S}{2}=1+\frac{1}{10}+\frac{1}{(10{)}^2}+\frac{1}{(10{)}^3}+...$

$\frac{S}{2}=\frac{1}{1-\frac{1}{10}}=\frac{10}{9}$ [infinite G.P sum $S=\frac{a}{1-r}]$

$S=\frac{20}{9}$

Option A is correct