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 an = a + (n – 1)d
∴ According to the given information,
$p^{\text {th }}$ term $=a_{p}=a+(p-1) d=\frac{1}{q}$ $\ldots(1)$
$q^{\text {th }}$ term $=a_{q}=a+(q-1) d=\frac{1}{p}$ $\ldots(2)$
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
$a+(p-1) \frac{1}{p q}=\frac{1}{q}$
$\Rightarrow a=\frac{1}{q}-\frac{1}{q}+\frac{1}{p q}=\frac{1}{p q}$
$\therefore S_{p q}=\frac{p q}{2}[2 a+(p q-1) d]$
$=\frac{p q}{2}\left[\frac{2}{p q}+(p q-1) \frac{1}{p q}\right]$
$=1+\frac{1}{2}(p q-1)$
$=\frac{1}{2} p q+1-\frac{1}{2}=\frac{1}{2} p q+\frac{1}{2}$
$=\frac{1}{2}(p q+1)$
Thus, the sum of first $p q$ terms of the A.P. is $\frac{1}{2}(p q+1)$.