Question

If $S$ is the sum to infinity of a GP, whose first term 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. none of these

Answer 1

Answer: (B)

$S[1-{(1-\frac{a}{S})}^n]$

Let r be the common ratio.
Given, $S_{\infty }=S=\frac{a}{1-r}$
$\Rightarrow 1-r=\frac{a}{S}$
$\Rightarrow r=1-\frac{a}{S}$
Now, sum of n terms is given by
$S_n=\frac{a(1-r^n)}{1-r}$

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

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

$=S[1-(1-\frac{a}{S}{)}^n]$
Option B is correct.