Let a, b, and c be the 7th , 11th and 13th terms respectively of
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:

1. $\frac{1}{2}$

2. 4

3. 2

4. $\frac{7}{13}$

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.

$\Rightarrow(\mathrm{A}+10 \mathrm{~d})^{2}=(\mathrm{A}+6 \mathrm{~d})(\mathrm{a}+12 \mathrm{~d})$

$\Rightarrow \frac{A}{d}=-14$

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