Question

Let $a,b$ and $c$ 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 $\frac{a}{c}$ is equal to :

• A. $4$
• B. $2$
• C. $\frac{7}{13}$
• D. $\frac{1}{2}$

Answer 1

The correct option is A $4$

Let $p$ and $d$ be the first term and common difference of A.P., respectively.
$7^{\text{th}},{11}^{\text{th}},{13}^{\text{th}}$ terms are in A. P.
$\Rightarrow p+6d,p+10d,p+12d$ are in A. P.

Also, $p+6d,p+10d,p+12d$ are in G. P.
Thus, $(p+10d{)}^2=(p+6d\left)\right(p+12d)$
$\Rightarrow 28d^2+2pd=0$
$\Rightarrow p=-14d$

$\therefore \frac{a}{c}=\frac{7^{\text{th}}\text{term}}{{13}^{\text{th}}\text{term}}$

$=\frac{p+6d}{p+12d}$

$=\frac{-14d+6d}{-14d+12d}$

$=4$