Question

If S_n = n² p and S_m = m² p, m ≠ n, in an A.P., prove that S_p = p³.

Answer 1

$$\begin{gathered} S_n=n^{2}p \\ \Rightarrow \frac{n}{2}\left[2a+(n-1)d\right]=n^{2}p \\ \Rightarrow 2np=2a+(n-1)d...\left(i\right) \\ {S}_m=m^{2}p \\ \Rightarrow \frac{m}{2}\left[2a+(m-1)d\right]=m^{2}p \\ \Rightarrow 2mp=2a+(m-1)d...\left(ii\right) \\ \mathrm{Subtracting}\left(\mathrm{ii}\right)\mathrm{from}\left(\mathrm{i}\right),\mathrm{we}\mathrm{get}: \\ 2p(n-m)=(n-m)d \\ \Rightarrow 2p=d...\left(iii\right) \\ \mathrm{Substituing}\mathrm{the}\mathrm{value}\mathrm{in}\left(\mathrm{i}\right),\mathrm{we}\mathrm{get}: \\ nd=2a+(n-1)d \\ \Rightarrow nd-nd+d=2a \\ \Rightarrow a=\frac{d}{2}=p\left[\mathrm{from}\left(\mathrm{iii}\right)\right]...\left(\mathrm{iv}\right) \\ \therefore S_p=\frac{p}{2}\left[2a+\left(p-1\right)d\right] \\ \Rightarrow S_p=\frac{p}{2}\left[2p+\left(p-1\right)2p\right] \\ \Rightarrow S_p=\frac{p}{2}\left[2p+2p^2-2p\right] \\ \Rightarrow S_p=\frac{p}{2}\left[2p^2\right] \\ \Rightarrow S_p=p^3 \end{gathered}$$