Question

If $S_n$ denotes the sum of $n$ terms of a $GP$ whose common ratio is $r$, then $\left(r-1\right) \frac{dS_{n}}{dr}$ 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
$\left(r-1\right) \frac{dS_{n}}{dr} + S_{n} = nar^{n-1} - 0$
$\Rightarrow \left(r-1\right) \frac{dS_{n}}{dr} = nar^{n-1} - S_{n}$
$= n$ [$n^{th}$ term of GP] $- S_{n}$
$= n\left[S_{n}-S_{n-1}\right]-S_{n}$
$= \left(n-1\right)S_{n}-n\,S_{n-1}$