Question

If $S_n=n^{2}p$ and $S_m=m^{2}p$, $m\ne n$, in A.P., then $S_p$ is?

• A. $p^2$
• B. $p^3$
• C. $p^4$
• D. None

Answer 1

The correct option is B $p^3$

$S_m=n^{2}p\cdot S_m=m^{2}p$

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

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

$\Rightarrow 2a+nd-d=2np-\left(1\right)$

$Sm=m^{2}p$

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

$\Rightarrow 2a+md-d=2mp-\left(2\right)$

subtract equation (2) from equation (1)

$(2a+nd-d)-(2a+md-d)=2np-2mp$

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

$\Rightarrow d=2p-\left(3\right)$

Put value of d from equation

(3) to equation (2)

2a+(m-1)(2p)=2 m p

$\Rightarrow a+mp-p=mp\Rightarrow a=p-\left(4\right)$

$\begin{array}{cc}\therefore Sp& =\frac{p}{2}\{2a+(p-1\left)d\right\}\\ & =\frac{p}{2}\{2p+(p-1\left)2p\right\}\\ Sp& =p^3\end{array}$

Answer: option

$\left(B\right)$