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

It is known that the general term of an $A.P.$ is $a_{n}=a+(n-1) d$

$\therefore$ According to the given information,

$p^{\text {th }}$ term $=a_{p}=a+(p-1) d=\frac{1}{q}$      ......$(1)$

$q^{ th }$ term $=a_{q}=a+(q-1) d=\frac{1}{p}$       ........$(2)$

Subtracting $(2)$ from $(1),$ we obtain

$(p-1) d-(q-1) d=\frac{1}{q}-\frac{1}{p}$

$\Rightarrow(p-1-q+1) d=\frac{p-q}{p q}$

$\Rightarrow(p-q) d=\frac{p-q}{p q}$

$\Rightarrow d=\frac{1}{p q}$

Putting the value of $d$ in $(1),$ we obtain $a+(p-1) \frac{1}{p q}=\frac{1}{q}$

$\Rightarrow a=\frac{1}{q}-\frac{1}{q}+\frac{1}{p q}=\frac{1}{p q}$

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

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

$=1+\frac{1}{2}(p q-1)$

$=\frac{1}{2} p q+1-\frac{1}{2}=\frac{1}{2} p q+\frac{1}{2}$

$=\frac{1}{2}(p q+1)$

Thus, the sum of first pq terms of the $A.P.$ is $=\frac{1}{2}(p q+1)$