Loopy AP

Let $a_1$ and $d$ be the first term and the common difference of the arithmetic progression respectively. Then

$\begin{cases} a_p = a_1 + (p-1)d = \dfrac 1q \\ a_q = a_1 + (q-1)d = \dfrac 1p \end{cases}$

$\begin{aligned} \implies a_p - a_q & = (p-q) d = \frac 1q - \frac 1p = \frac {p-q}{pq} \\ \implies d & = \frac 1{pq} \\ \implies a_p & = a_1 + \frac {p-1}{pq} = \frac 1q \\ \implies a_1 & = \frac 1{pq} \end{aligned}$

Then $a_{pq} = a_1 + (pq-1) d = \dfrac 1{pq} + \dfrac {pq-1}{pq} = \boxed 1$ .