Question

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

Answer 1

\[S_n = n^2 p\]

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

\[ \Rightarrow 2np = 2a + (n - 1)d . . . (i)\]

\[ S_m = m^2 p\]

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

\[ \Rightarrow 2mp = 2a + (m - 1)d . . . (ii)\]

\[\text { Subtracting (ii) from (i), we get }: \]

\[2p(n - m) = (n - m)d\]

\[ \Rightarrow 2p = d . . . (iii)\]

\[\text { Substituing the value in (i), we get }: \]

\[nd = 2a + (n - 1)d\]

\[ \Rightarrow nd - nd + d = 2a\]

\[ \Rightarrow a = \frac{d}{2} = p \left[ \text { from }(iii) \right] . . . (iv)\]

\[ \therefore S_p = \frac{p}{2}\left[ 2a + \left( p - 1 \right)d \right]\]

\[ \Rightarrow S_p = \frac{p}{2}\left[ 2p + \left( p - 1 \right)2p \right]\]

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

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

\[ \Rightarrow S_p = p^3 \]