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)

$S = \frac{4}{19} +\frac{44}{19^{2} } +\frac{444}{19^{3}} + ....\infty \quad...\left(i\right)$
$ \Rightarrow \frac{S}{19}= \frac{4}{19^{2}} +\frac{44}{19^{3}} + .... \infty\quad...\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}$