Question

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

• A. $S{(1-\frac{a}{S})}^n$
• B. $S{[1-(1-\frac{a}{S})]}^n$
• C. $a{[1-(1-\frac{a}{S})]}^n$
• D. $S{[1-(1-\frac{S}{a})]}^n$

Answer 1

The correct option is A $S{(1-\frac{a}{S})}^n$

$Formulaused$
$S_{\propto }=\frac{a}{1-r}$
$S_n=\frac{a(1-r^n)}{1-r}$
$L\text{et the given G}\text{.P series be}$
$\text{a,ar,a}{\text{r}}^2,a{r}^3,........\propto$
$S=\frac{a}{1-r}$
$\Rightarrow 1-r=\frac{a}{S}$
$\Rightarrow r=1-\frac{a}{S}...........\left(i\right)$
$S_n=\frac{a(1-r^n)}{1-r}$
$=\frac{a(1-{(1-\frac{a}{S})}^n)}{1-(1-\frac{a}{S})}from\left(i\right)$
$=\frac{a(1-{(1-\frac{a}{S})}^n)}{\left(\frac{a}{S}\right)}$
$=S(1-{(1-\frac{a}{S})}^n)$