Common ratio of a GP - $a_p, a_q, a_r$ given $a_1, a_2 ...$ form an AP

Question

Given $a_1, a_2 ...$ form an AP and $a_p, a_q, a_r$ form a GP. Find $\dfrac{a_q}{a_r}$

I did :

$$a_p, a_q, a_r \text{ form a GP}$$

$$\implies a_q^2 = a_p\cdot a_r$$

I got till here :

$$a(2q-p-r) = d(2q-p-r+pr-q^2)$$

I cant eleminate a and d. The answer given was in ratio of p,q,r or:

Answer: $\dfrac{r-q}{q-p}$

Thank you.

Answer 1

Since \begin{align*} a_q^2 & = a_pa_r\\ a_q^2 -a_ra_q& = a_pa_r-a_ra_q\\ a_q(a_q-a_r)&=a_r(a_p-a_q)\\ \frac{a_q}{a_r}& =\frac{a_p-a_q}{a_q-a_r}\\ \frac{a_q}{a_r}& =\frac{d(p-q)}{d(q-r)}\\ \frac{a_q}{a_r}& =\frac{p-q}{q-r}\\ \end{align*} My answer and yours doesn't match (they are close in form). Perhaps check your answer again. If you want the common ratio then it will be $\dfrac{a_r}{a_q}=\frac{q-r}{p-q}$.