In an $A.P.$,if the $p^{\text{th}}$ term is $\frac{1}{q}$ and the $q^{\text{th}}$ term is $\frac{1}{p}$,prove that the sum of the first $pq$ terms is $\frac{1}{2}(pq+1)$,where $p \neq q$.

The general term of an $A.P.$ is $a_n = a + (n-1)d$.
According to the given information:
$a_p = a + (p-1)d = \frac{1}{q}$ $(1)$
$a_q = a + (q-1)d = \frac{1}{p}$ $(2)$
Subtracting $(2)$ from $(1)$:
$(p-1)d - (q-1)d = \frac{1}{q} - \frac{1}{p}$
$(p-q)d = \frac{p-q}{pq}$
Since $p \neq q$,we have $d = \frac{1}{pq}$.
Substituting $d$ into $(1)$:
$a + (p-1)\frac{1}{pq} = \frac{1}{q}$
$a = \frac{1}{q} - \frac{p-1}{pq} = \frac{p - (p-1)}{pq} = \frac{1}{pq}$.
The sum of the first $pq$ terms is $S_{pq} = \frac{pq}{2}[2a + (pq-1)d]$.
$S_{pq} = \frac{pq}{2}[2(\frac{1}{pq}) + (pq-1)(\frac{1}{pq})]$
$S_{pq} = \frac{pq}{2}[\frac{2 + pq - 1}{pq}]$
$S_{pq} = \frac{pq}{2}[\frac{pq+1}{pq}] = \frac{1}{2}(pq+1)$.