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
( $\because$ The given $G.P$. is a decreasing one)