Question

Three numbers, the third of which being $12$, form a decreasing $G. P$. If the third term were $9$ instead of $12$, the three numbers would have formed an $A.P$. Then the common ratio of the original $G.P$., is

• A. $\frac{3}{7}$
• B. $\frac{2}{3}$
• C. $\frac{3}{5}$
• D. $\frac{4}{5}$

Answer 1

Answer: (B)

Let $ \frac{12}{r^{2}}, \frac{12}{r}, 12$ be a decreasing $G.P$.
$ \therefore $ By data $\frac{12}{r^{2}}, \frac{12}{r}, 9$ are in $A.P$.
$ \Rightarrow \frac{24}{r} = 9 + \frac{12}{r^{2}}$
$\Rightarrow 3r^{2} -8r +4 = 0 $
$\Rightarrow r= 2$ or $\frac{2}{3}$
$ \Rightarrow r= \frac{2}{3}$ only
$ \quad\quad\quad $ ( $\because$ The given $G. P$. is a decreasing one)