Question

The next term of the series $\sqrt{3},\sqrt{12},\sqrt{27}$ is:

• A. $\sqrt{39}$
• B. $\sqrt{32}$
• C. $\sqrt{54}$
• D. $\sqrt{48}$

Answer 1

The correct option is D $\sqrt{48}$

Given $\sqrt{3},\sqrt{12},\sqrt{27}$

$\Rightarrow \sqrt{3},\sqrt{4\times 3},\sqrt{9\times 3}$

$\Rightarrow \sqrt{3},2\sqrt{3},3\sqrt{3}$ are in A.P.

where $a=\sqrt{3}$ and $d=2\sqrt{3}-\sqrt{3}=\sqrt{3}$

$\therefore$ the fourth term is $T_4=a+3d=\sqrt{3}+3\times \sqrt{3}=4\sqrt{3}=\sqrt{16\times 3}=\sqrt{48}$