Question

Let $S=\frac{4}{19}+\frac{44}{19^2}+\frac{444}{19^3}+.....\infty$. Then $S$ is equal to

• A. $\frac{38}{81}$
• B. $\frac{4}{19}$
• C. $\frac{36}{171}$
• D. $\frac{1}{19}$

Answer 1

Answer: (A)

$S=\frac{4}{19}+\frac{44}{19^2}+\frac{444}{19^3}+....\infty ...\left(i\right)$
$\Rightarrow \frac{S}{19}=\frac{4}{19^2}+\frac{44}{19^3}+....\infty ...\left(ii\right)$
Subtracting $\left(ii\right)$ from $\left(i\right)$, we get
$S\frac{18}{19}=\frac{4}{19}+\frac{40}{19^2}+\frac{400}{19^3}+.....\infty$
$=\frac{4}{19}\left[1+\frac{10}{19}+{\left(\frac{10}{19}\right)}^2+.....\infty \right]$
$=\frac{4}{19}\left[\frac{1}{1-\frac{10}{19}}\right]=\frac{4}{9}$
$\Rightarrow S=\frac{76}{162}$
$=\frac{38}{81}$