Question

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

Answer 1

Given that

$\Rightarrow S_n=n^{2}p$

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

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

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

Also given that

$\Rightarrow S_m=m^{2}p$

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

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

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

Now, $\left(1\right)-\left(2\right)$

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

$\Rightarrow nd-d-md+d=2pn-2pm$

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

$\Rightarrow d=2p$

Substitute this in $\left(1\right)$, we get

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

$\Rightarrow 2a+2np-2p=2np$

$\Rightarrow 2a=2p$

$\Rightarrow a=p$

Now, consider $S_p$

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

Substitute the values of $a,d$ in the above equation, we get

$\Rightarrow S_p=\frac{p}{2}(2p+(p-1\left)2p\right)$

$\Rightarrow S_p=p(p+(p-1\left)p\right)$

$\Rightarrow S_p=p(p+p^2-p)$

$\Rightarrow S_p=p\left(p^2\right)$

$\Rightarrow S_p=p^3$

Hence proved.