Question

If $S_n$ denotes the sum of $n$ terms of a $GP$ whose common ratio is $r$, then $\left(r-1\right)\frac{d{S}_n}{dr}$ is equal to

• A. $\left(n-1\right)S_n+n{S}_{n-1}$
• B. $\left(n-1\right)S_n-n{S}_{n-1}$
• C. $\left(n-1\right)S_n$
• D. None of these

Answer 1

Answer: (B)

We have, $S_n=\frac{a\left(r^n-1\right)}{r-1}$
$\Rightarrow \left(r-1\right)S_n=a{r}^n-a$
On differentiating both sides w.r.t r, we get
$\left(r-1\right)\frac{d{S}_n}{dr}+S_n=na{r}^{n-1}-0$
$\Rightarrow \left(r-1\right)\frac{d{S}_n}{dr}=na{r}^{n-1}-S_n$
$=n$ [$n^{th}$ term of GP] $-S_n$
$=n\left[S_n-S_{n-1}\right]-S_n$
$=\left(n-1\right)S_n-n{S}_{n-1}$