The $p^{th}, q^{th},r^{th}$ term of an A.P are in G.P . Find the ratio of the $p^{th}$ and the $r^{th}$ term. I took the $p^{th}, q^{th},r^{th}$ term as $t_p,t_q,t_r$ respectively. Then $t_r = t_p * c^2$ (If c is the common ratio). and the common difference of the A.P is $\frac{t_r -t_p}{r-p}$. But I think I need to relate the common difference and the common ratio.
 A: Hint:
If  $a,d(\ne0)$ be the first term & the common difference of the AP
$$\{a+(p-1)d\}\{a+(r-1)d\}=\{a+(q-1)d\}^2$$
$a^2+ad(p+r-2)+(r-1)(p-1)d^2=a^2+2ad(q-1)+(q-1)^2d^2$
As $d\ne0$
$$a(p+r-2)+(r-1)(p-1)d=2a(q-1)+(q-1)^2d$$
Express $a$ in terms of $d$
Now we need $$\dfrac{a+(p-1)d}{a+(r-1)d}$$
Replace the value of $a$ in terms of $d$
A: Let $a,b,c$ be the  $p^{\text{th}}, q^{\text{th}},r^{\text{th}}$ terms  of the AP respectively, and let $\alpha=\frac ac, \beta=\frac bc$.
As $a,b,c$ are consecutive terms of a GP, 
$$\begin{align}
b^2&=ac\\
\left(\frac bc\right)^2&=\frac ac\\
\Longrightarrow \beta^2&=\alpha\end{align}$$
As $a,b,c$ are the $p^{\text{th}}, q^{\text{th}},r^{\text{th}}$ terms of an AP, 
$$\begin{align}
\frac {q-p}{r-q}&=\frac {b-a}{c-b}\\
&=
\frac {\beta-\alpha}{1-\beta}\\
&=\frac {\beta-\beta^2}{1-\beta}
&&\scriptsize (\because \alpha=\beta^2)\\
&=\beta\\
\therefore \alpha=\frac ac&=\color{red}{\left(\frac {q-p}{r-q}\right)^2}
&&\scriptsize (\because \alpha=\beta^2)\\\end{align}$$
