Question

Insert 3 Geometric means between 1 and 9.

• A. 3
• B. $3,9,27$
• C. $3\sqrt{3},3,\sqrt{3}$
• D. $3\sqrt{3}6\sqrt{3}$

Answer 1

The correct option is A 3

Let G1, G2 and G3 be the three geometric means. Thus 1, G1, G2, G3, 9 is the required G.P

a = 1, $a{r}^4=9$

$\therefore 1\times r^4=9$ or $(r^2{)}^2=(3{)}^2$

$\Rightarrow r=\sqrt{3}$

$G1=ar=1\times \sqrt{3}=\sqrt{3}$

$G2=a{r}^2=1\times (\sqrt{3}{)}^2=3$

$G3=a{r}^3=1\times (\sqrt{3}{)}^2.\sqrt{3}=3\sqrt{3}$

$\therefore$ The 3 G.Ms are $\sqrt{3},3,3\sqrt{3}$