Question

If $4$ GM's be inserted between $160$ and $5$, then third GM will be-

• A. 8
• B. 118
• C. 20
• D. 40

Answer 1

Answer: (C)

20

The G.P is

$160,\underset{–}{G_1},\underset{–}{G_2},\underset{–}{G_3},\underset{–}{G_4},5$

First term$=a=160$

Let the $4$ G.M. be $G_1,G_2,G_3,G_4$

$\therefore G_5=$ last term $=5$

$\therefore G_5=a_6=a{r}^{6-1}$

$=a{r}^5$

$\therefore a{r}^5=5$

$\Rightarrow 160r^5=5$

$\Rightarrow r^5=\frac{5}{160}$

$\Rightarrow r={\left(\frac{5}{160}\right)}^{1/5}={\left(\frac{1}{32}\right)}^{1/5}={\left(\frac{1}{2^5}\right)}^{1/5}={\left(\frac{1}{2}\right)}^{5\times 1/5}$

$\Rightarrow r=\frac{1}{2}$

$\therefore r=\frac{1}{2}$

$\therefore G_3=a_4=a{r}^{4-1}=a{r}^3$

$=160\times {\left(\frac{1}{2}\right)}^3=\frac{160}{8}=20$

$\therefore G_3=20$

option $C$.