If pth, qth and rth terms of an A.P. are in G.P., then the common ratio of this G.P. is

(b) $\frac{q-r}{p-q}$

Let a be the first term and d be the common difference of the given A.P.

Then, we have:

$\mathrm{p}^{\text {th }}$ term, $a_{p}=a+(p-1) d$

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

$r^{\text {th }}$ term, $a_{r}=a+(r-1) d$

Now, according to the question the $p^{\text {th }}$, the $q^{\text {th }}$ and the $r^{\text {th }}$ terms are in G.P.

$\therefore(a+(q-1) d)^{2}=(a+(p-1) d) \times(a+(r-1) d)$

$\Rightarrow a^{2}+2 a(q-1) d+((q-1) d)^{2}=a^{2}+a d(r-1+p-1)+(p-1)(r-1) d^{2}$

$\Rightarrow a d(2 q-2-r-p+2)+d^{2}\left(q^{2}-2 q+1-p r+p+r-1\right)=0$

$\Rightarrow a(2 q-r-p)+d\left(q^{2}-2 q-p r+p+r\right)=0 \quad(\because d$ cannot be 0$)$

$\Rightarrow a=-\frac{\left(q^{2}-2 q-p r+p+r\right) d}{(2 q-r-p)}$

$\therefore$ Common ratio, $r=\frac{a_{q}}{a_{p}}$

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

$=\frac{\frac{\left(q^{2}-2 q-p r+p+r\right) d}{(p+r-2 q)}+(q-1) d}{\frac{\left(q^{2}-2 q-p r+p+r\right) d}{(p+r-2 q)}+(p-1) d}$

$=\frac{q^{2}-2 q-p r+p+r+p q+r q-2 q^{2}-p-r+2 q}{q^{2}-2 q-p r+p+r+p^{2}+p r-2 p q-p-r+2 q}$

$=\frac{p q-p r-q^{2}+q r}{p^{2}+q^{2}-2 p q}$

$=\frac{p(q-r)-q(q-r)}{(p-q)^{2}}$

$=\frac{(p-q)(q-r)}{(p-q)^{2}}$

$=\frac{(q-r)}{(p-q)}$