Question:
If $S_{m}=m^{2} p$ and $S_{n}=n^{2} p$, where $m \neq n$ in an AP then prove that $S_{p}=p^{3}$.
Solution:
Let the first term of the AP be a and the common difference be d
Given: $S_{m}=m^{2} p$ and $S_{n}=n^{2} p$
To prove: $\mathrm{Sp}=\mathrm{p}^{3}$
According to the problem
$\frac{m}{2}[2 a+(m-1) d]=m^{2} p \Rightarrow 2 a+(m-1) d=2 m p$
and $\frac{n}{2}[2 a+(n-1) d]=n^{2} p \Rightarrow 2 a+(n-1) d=2 n p$
Subtracting the equations we get,
$(m-n) d=2 p(m-n)$
Now m is not equal to n
So $d=2 p$
Substituting in $1^{\text {st }}$ equation we get
$2 a+(m-1)(2 p)=2 m p$
$\Rightarrow a=m p-m p+p=p$
$\Rightarrow S_{p}=\frac{p}{2}[2 p+(p-1)(2 p)]$
$\Rightarrow S_{p}=\frac{p}{2}\left[2 p+2 p^{2}-2 p\right]=p^{3}$
Hence proved.