Question

If the sum of $m$ terms of an A.P. is the same as the sum of its $n$ terms, show that the sum of its $(m+n)$ terms is zero.

Answer 1

Let $a$ be the first term and $d$ be the common difference of the given A.P. Then, $S_m=S_n$.
$⟹\frac{m}{2}\{2a+(m-1\left)d\right\}=\frac{n}{2}\{2a+(n-1\left)d\right\}$

$⟹2a(m-n)+\left\{m\right(m-1)-n(n-1\left)\right\}d=0$

$⟹2a(m-n)+\left\{\right(m^2-n^2)-(m-n\left)\right\}d=0$

$⟹(m-n)\{2a+(m+n-1\left)d\right\}=0$

$⟹2a+(m+n-1)d=0$ [$\because m-n\ne 0$] ...(i)
Now,
$S_{m+n}=\frac{m+n}{2}\{2a+(m+n-1\left)d\right\}$

$⟹S_{m+n}=\frac{m+n}{2}\times 0=0$ [Using (i)]