If in an A.P. Sn = n2p and Sm = m2p, where Sr denotes the sum of r terms of the A.P., then Sp is equal to

In the given problem, we are given an A.P whose $S_{n}=n^{2} \mathrm{p}$ and $S_{n}=m^{2} \mathrm{p}$

We need to find $S_{p}$

Now, as we know,

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

Where, first term = a

Common difference = d

Number of terms = n

So,

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

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

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

Similarly,

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

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

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

Equating (1) and (2), we get,

$\frac{1}{2 n}(2 a+n d-d)=\frac{1}{2 m}(2 a+m d-d)$

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

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

Solving further, we get,

$2 a m-2 a n=-n d+m d$

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

$2 a=d$ ...(3)

Further, substituting (3) in (1), we get,

$S_{s}=\frac{n}{2}[d+(n-1) d]$

$n^{2} p=\frac{n}{2}[d+n d-d]$

$p=\frac{1}{2 n}[n d]$

$p=\frac{d}{2}$ ......(4)

Now,

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

$S_{p}=\frac{p}{2}[d+p d-d]$(Using 3)

$S_{D}=\frac{p}{2}[p(2 p)]$(Using 4)

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

Thus, $S_{p}=p^{3}$

Therefore, the correct option is (C).