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 + n) p2
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
a = p
Sum of p terms of the A.P. is given by:
$\frac{p}{2}\{2 a+(p-1) d\}$
$=\frac{p}{2}\{2 p+(p-1) 2 p\}$
$=p^{3}$