Question

Let $a, b$ and $c$ be the $7^{th}, 11^{th}$ and $13^{th}$ 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. $\frac{1}{2}$
• B. $4$
• C. $2$
• D. $\frac{7}{13}$

Answer 1

Answer: (B)

$a = A + 6d$
$b = A + 10d$
$c = A + 12d$
a,b,c are in G.P.
$\Rightarrow \; (A + 10d)^2 = (A 6d)(a + 12d)$
$\Rightarrow \; \frac{A}{d} = - 14 $
$ \frac{a}{c} = \frac{A + 6d}{A + 12d} = \frac{ 6 + \frac{A}{d}}{12 + \frac{A}{d}} = \frac{6 - 14}{12 -14} = 4 $