If mth term of an AP is

Solution:

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,

$S_{m n}=\frac{m n}{2}[2 a+(m n-1) d]$

$=\frac{m n}{2}\left[\frac{2}{m n}+(m n-1) \frac{1}{m n}\right]$

$=\frac{m n}{2}\left(\frac{1+m n}{m n}\right)$

$=\left(\frac{m n+1}{2}\right)$