If $m$ th term of an AP is $\frac{1}{n}$ and $n$th term is $\frac{1}{m}$ then find the sum of its first $m n$ terms.
Suppose a be the first term and d be the common difference of the given AP.
$a_{m}=\frac{1}{n}$
$\Rightarrow a+(m-1) d=\frac{1}{n} \quad \ldots \ldots(1)$
And,
$a_{n}=\frac{1}{m}$
$\Rightarrow a+(n-1) d=\frac{1}{m} \quad \ldots .(2)$
Subtracting (2) from (1), we get
$\frac{1}{n}-\frac{1}{m}=(m-n) d$
$\Rightarrow \frac{m-n}{m n}=(m-n) d$
$\Rightarrow d=\frac{1}{m n}$
Putting $d=\frac{1}{m n}$ in (1), we get
$a+(m-1) \frac{1}{m n}=\frac{1}{n}$
$\Rightarrow a+\frac{1}{n}-\frac{1}{m n}=\frac{1}{n}$
$\Rightarrow a=\frac{1}{m n}$
∴ Sum of mn terms,
$S_{m n}=\frac{m n}{2}[2 a+(m n-1) d]$
$=\frac{m n}{2}\left[\frac{2}{m n}+(m n-1) \frac{1}{m n}\right]$
$=\frac{m n}{2}\left(\frac{1+m n}{m n}\right)$
$=\left(\frac{m n+1}{2}\right)$