Let $a, b and c$be the 7th' 11th and 13th terms respectively of a non-constant A.P, If these are also

the three consecutive terms of a G.P., then $\frac{a}{c}=$

We have, $a=A+6d;b=A+10d;c=A+12d$

Since, $a, b$and $c$ are in G.P .

$\begin{array}{l}\therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}(A+10d{)}^2=(A+6d\left)\right(A+12d)\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{A}^2+100{d^2}^2+20Ad=A^2+18Ad+72d^2\end{array}$

$\Rightarrow 2Ad=-28d^2\Rightarrow \frac{A}{d}=-14$

$[\because d\ne 0$ as A.P. is non constant]

Now, $\frac{a}{c}=\frac{A+6d}{A+12d}=\frac{\frac{A}{d}+6}{\frac{A}{d}+12}=\frac{-14+6}{-14+12}=4$