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nth Term of A.P

If sn be th...

Question

If $s_{n}$ be the sum of n terms of an A.P. and if $\frac{s_{p n}}{s_{n}}$ is independent of n, then prove that $d = 2 a$ in usual notation.

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Solution

$\frac{s_{p n}}{s_{n}} = \frac{\frac{p n}{2} [2 a + (p n - 1) d]}{\frac{n}{2} [2 a + (n - 1) d]} = \frac{p [2 a - d + p n d]}{[2 a - d + n d]}$
Above will be independent of n if $2 a - d = 0$ and in that case the ratio will be $p^{2}$ as nd cancels.

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