# Solve the following

Question:

If in an A.P., Sn = n2p and Sm = m2p, where Sr denotes the sum of r terms of the A.P., then Sp is equal to

(a) $\frac{1}{2} p^{3}$

(b) mn p

(c) P3

(d) (m + np2

Solution:

(c) $P^{3}$

Given:

$S_{n}=n^{2} p$

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

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

$\Rightarrow 2 a=2 n p-(n-1) d$    ....(1)

$S_{m}=m^{2} p$

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

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

$\Rightarrow 2 a=2 m p-(m-1) d$    ...(2)

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

$2 n p-(n-1) d=2 m p-(m-1) d$

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

$\Rightarrow 2 p=d$

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

a = p

Sum of terms of the A.P. is given by:

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

$=\frac{p}{2}\{2 p+(p-1) 2 p\}$

$=p^{3}$