Solve the following

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

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

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

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

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

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

Subtracting (ii) from (i), we get:

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

$\Rightarrow 2 p=d \quad \ldots(i i i)$

Substituing the value in (i), we get:

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

$\Rightarrow n d-n d+d=2 a$

$\Rightarrow a=\frac{d}{2}=p[$ from(iii) $] \quad \ldots$ (iv)

$\therefore S_{p}=\frac{p}{2}[2 a+(p-1) d]$

$\Rightarrow S_{p}=\frac{p}{2}[2 p+(p-1) 2 p]$

$\Rightarrow S_{p}=\frac{p}{2}\left[2 p+2 p^{2}-2 p\right]$

$\Rightarrow S_{p}=\frac{p}{2}\left[2 p^{2}\right]$

$\Rightarrow S_{p}=p^{3}$