Question:
Let $\mathrm{a}, \mathrm{b}$ and $\mathrm{c}$ be the $7^{\text {th }}, 11^{\text {th }}$ and $13^{\text {th }}$ terms respectively of a non-constant A.P. If these are
also the three consecutive terms of a G.P., then $\frac{\mathrm{a}}{\mathrm{c}}$ is equal to:
Correct Option: , 2
Solution:
$a=A+6 d$
$b=A+10 d$
$c=A+12 d$
$\mathrm{a}, \mathrm{b}, \mathrm{c}$ are in G.P.
$\Rightarrow(\mathrm{A}+10 \mathrm{~d})^{2}=(\mathrm{A}+6 \mathrm{~d})(\mathrm{a}+12 \mathrm{~d})$
$\Rightarrow \frac{A}{d}=-14$
$\frac{\mathrm{a}}{\mathrm{c}}=\frac{\mathrm{A}+6 \mathrm{~d}}{\mathrm{~A}+12 \mathrm{~d}}=\frac{6+\frac{\mathrm{A}}{\mathrm{d}}}{12+\frac{\mathrm{A}}{\mathrm{d}}}=\frac{6-14}{12-14}=4$