If pth, qth, rth and sth terms of an A.P. be in G.P.,

Here, $a_{p}=a+(p-1) d$

$a_{q}=a+(q-1) d$

$a_{r}=a+(r-1) d$

$a_{s}=a+(s-1) d$

It is given that $a_{p}, a_{q}, a_{r}$ and $a_{s}$ are in G.P.

$\therefore \frac{a_{q}}{a_{p}}=\frac{a_{r}}{a_{q}}=\frac{a_{q}-a_{r}}{a_{p}-a_{q}}=\frac{q-r}{p-q} \quad \ldots \ldots(\mathrm{i})$

Similarly, $\frac{a_{r}}{a_{q}}=\frac{a_{s}}{a_{r}}=\frac{a_{r}-a_{s}}{a_{q}-a_{r}}=\frac{r-s}{q-r} \quad \ldots \ldots$ (ii)'

Using (i) and (ii) :

$\frac{q-r}{p-q}=\frac{r-s}{q-r}$

Therefore, $p-q, q-r$ and $r-s$ are in G. P.