Let $T_r$ be the rth term of an AP, for $r=1,$ 2…. . If for some positive integers m and n, we have $T_m=\frac{1}{n}$ and

 $T_n=\frac{1}{m},$ the $T_{m+n}$ equal 

Let a be the first term and d be the com-mon difference of the given A.P. Then according to the

hypothesis,

$\begin{array}{l} T_m-T_n=\frac{1}{n}-\frac{1}{m}\\ \Rightarrow (m-n)d=\frac{m-n}{mn}\Rightarrow d=\frac{1}{mn}.\\ \Rightarrow T_{m+n}-T_m=(m+n-m)\frac{1}{mn}=\frac{1}{m}\end{array}$