If the sum of  $m$ terms of an A.P is the same as the sum of its $n$  terms, then the sum of its  $\left(m+n\right)$ term is

 Let a be the first term and d be the common differences of the given A.P, then 

$\begin{array}{l}{S}_m=S_n\Rightarrow \frac{m}{2}[2a+(m-1\left)d\right]=\frac{n}{2}[2a+(n-1\left)d\right]\\ \Rightarrow 2a(m-n)+\left\{m\right(m-1)-n(n-1\left)\right\}d=0\\ \Rightarrow 2a(m-n)+\left\{\left(m^2-n^2\right)-(m-n)\right\}d=0\\ \Rightarrow (m-n)[2a+(m+n-1\left)d\right]=0\\ \Rightarrow 2a+(m+n-1)d=0[\because m-n\ne 0]\\ \text{ Now, }\\ S_{m+n}=\frac{m+n}{2}[2a+(m+n-1\left)d\right]=\frac{m+n}{2}\times 0=0\end{array}$