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If $S_{m}=m^{2} p$ and $S_{n}=n^{2} p$, where $m \neq n$ in an AP then prove that $S_{p}=p^{3}$.



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.


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