Question

If in an A.P., $S_n=n^{2}p$ and $S_m=m^{2}p$, where $S_r$ denotes the sum of r terms of the A.P., then $S_p$ is equal to

• A. $\frac{1}{2}{p}^3$
• B. $mnp$
• C. $p^3$
• D. $(m+n)p^2$

Answer 1

The correct option is C

$p^3$

Given:

$S_n=n^{2}p$

$\Rightarrow \frac{n}{2}\{2a+(n-1\left)d\right\}=n^{2}p$

$\Rightarrow 2a+(n-1)d=2np$

$\Rightarrow 2a=2np-(n-1)d\dots \dots \left(1\right)$

$S_m=m^{2}p$

$\Rightarrow \frac{m}{2}\{2a+(m-1\left)d\right\}=m^{2}p$

$\Rightarrow 2a+(m-1)d=2mp$

$\Rightarrow 2a=2mp-(m-1)d=2mp\dots \dots \left(2\right)$

From (1) and (2), we have:

2np - (n - 1)d = 2mp - (m - 1)d

$\Rightarrow 2p(n-2)=d(n-1-m+1)$

$\Rightarrow 2p=d$

Substituting d = 2p in equation (1), we get:

a = p

Sum of p terms of the A.P. is given by:

$\frac{p}{2}\{2a+(p-1\left)d\right\}$

$=\frac{p}{2}\{2p+(p-1\left)2p\right\}=p^3$