If the $m^{th}$   term of an $H.P$  is $\text{'}n\text{'}$  and nth term is $\text{'}m\text{'}$ , then its ${\left(mn\right)}^{th}$  term is

$a_m=n,a_n=m$

$\frac{1}{a+\left(m-1\right)d}=n,\frac{1}{a\left(n-1\right)d}=m$

$\begin{array}{c}a+\left(m-1\right)d=\frac{1}{n}\to \left(1\right)\\ a+\left(n-1\right)d=\frac{1}{m}\to \left(2\right)\\ ---\\ \_\_\_\_\_\_\_\_\_\_\_\_\_\\ \left(m-n\right)d=\frac{m-n}{mn}\end{array}$

$d=\frac{1}{mn}$

Substitute $d=\frac{1}{mn}$  in eqn(1)

$a+\frac{m-1}{mn}=\frac{1}{n}$

$a=\frac{1}{n}-\frac{m-1}{mn}$

$=\frac{m-m+1}{mn}=\frac{1}{mn}$

$a_{mn}=a+\left(mn-1\right)d$

$=\frac{1}{mn}+\left(mn-1\right)\frac{1}{mn}$

$a_{mn}=1$

$a_{mn}$  in $H.P=1$