Question

If $S$ is the sum to infinite of a G.P., whose first term is $a$, then the sum of the first $n$ terms is

• A. $S{\left(1-\frac{a}{S}\right)}^n$
• B. $S\left(1-{\left(1-\frac{a}{S}\right)}^n\right)$
• C. $a\left(1-{\left(1-\frac{a}{S}\right)}^n\right)$
• D. None of these

Answer 1

The correct option is B

$S\left(1-{\left(1-\frac{a}{S}\right)}^n\right)$

Explanation for the correct option.

Given, $S$ is the sum to infinity of a G.P., whose first term is $a$.

We know that the sum of infinite terms in GP series is given by:

$S=\frac{\mathrm{a}}{(1-\mathrm{r})}$
$\begin{array}{rcl}& \Rightarrow & \mathrm{r}=1-\frac{\mathrm{a}}{\mathrm{S}}\\ & \Rightarrow & {\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{a}\left(1-{\mathrm{r}}^{\mathrm{n}}\right)}{(1-\mathrm{r})}\\ & \Rightarrow & {\mathrm{S}}_{\mathrm{n}}=\mathrm{S}\left(1-{\mathrm{r}}^{\mathrm{n}}\right)\\ & \Rightarrow & {\mathrm{S}}_{\mathrm{n}}=\mathrm{S}\left(1-{\left(1-\frac{\mathrm{a}}{\mathrm{S}}\right)}^{\mathrm{n}}\right)\left[\because \mathrm{r}=\left(1-\frac{\mathrm{a}}{\mathrm{S}}\right)\right]\end{array}$

Hence, option B is correct.