$\text{ If }p\text{ th, }q\text{ th and }r\text{ th terms of an A.P. are in G.P. whose common ratio is }k,\text{ then }$

$\mathrm{the} \mathrm{root} \mathrm{of} \mathrm{equation} (q-r)x^2+(r-p)x+(p-q)=0\text{ other than unity is }$

$If \frac{x}{y}=\frac{y}{z}=k then \frac{x}{y}=\frac{y}{z}=\frac{x-y}{y-z}=k$

$\begin{array}{l}k=\frac{a+(q-1)d}{a+(p-1)d}=\frac{a+(r-1)d}{a+(q-1)d}\\ k=\frac{a+(q-1)d-a-(r-1)d}{a+(p-1)d-a-(q-1)d}\\ k=\frac{(q-r)d}{(p-q)d}=\frac{(q-r)}{p-q}\\ from the equation product of one root x1=\frac{p-q}{q-r}=\frac{1}{k}\end{array}$

$\text{ Since one root is }1,\text{ therefore other root }=\frac{1}{k}=\frac{p-q}{q-r}$