Question

If $S_n=n^{2}p$ and $S_m=m^{2}p$, $m\ne n$ in an A.P., prove that $S_n=p^3$.

Answer 1

Given,

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

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

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

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

(1)-(2) gives,

$2np-2mp=2a+(n-1)d-2a+(m-1)d$

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

$\therefore d=2p$

From (1)

$2a+(n-1)\left(2p\right)=2np$

$a+(n-1)p=np$

$\therefore a=p$

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

$=\frac{p}{2}\left[2\right(p)+(p-1\left)2p\right]$

$=p[p+p^2-p]$

$=p^3$

Hence proved.