Question

If five $G.M{.}^{\prime }s$ are inserted between $486$ and $\frac{2}{3}$ then fourth $G.M$ will be

• A. $4$
• B. $6$
• C. $12$
• D. $-6$

Answer 1

Answer: (B)

$6$

Let $G_{1,}{G}_{2,}{G}_{3,}{G}_{4,}{G}_5$ be five G.M's

Then $486$,$G_{1,}{G}_{2,}{G}_{3,}{G}_{4,}{G}_5$$,\frac{2}{3}$ is a G.P with $a=486$ and $a_7=\frac{2}{3}$

Now $a_7=\frac{2}{3}$

$\Rightarrow a{r}^6=\frac{2}{3}$

$\Rightarrow 486r^6=\frac{2}{3}$

$\Rightarrow r^6=\frac{2}{3}\times \frac{1}{486}$

$\Rightarrow r^6=\frac{1}{3\times 243}$

$\Rightarrow r^6=\frac{1}{3^6}$

$\Rightarrow r=\pm \frac{1}{3}$

Now, $G_4=a_5=a{r}^4=486\times {\left(\frac{1}{3}\right)}^4=\frac{486}{81}=6$