If m times the $mth$ term of an A.P. with non-zero common difference equals $n$ times the $nth$ term of the A.P., where $m\ne n$, then $(m+n)th$ term of this A.P. is 

Let $a$ be the first term and $d$ be the common difference of the A.P. We thve
$\begin{array}{l} \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}m{a}_m=n{a}_n\\ \Rightarrow \hspace{0.25em} \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}m[a+(m-1\left)d\right]=n[a+(n-1\left)d\right]\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em} (m-n)(a-d)+\left(m^2-n^2\right)d=0\end{array}$
$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}(m-n)[a-d+(m+n\left)d\right]=0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}a+(m+n-1)d=0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{a}_{m+n}=0\end{array}[\because m\ne \mathrm{n}]$