Question

The sum of $10$ terms of the series $\sqrt{3}+\sqrt{12}+\sqrt{27}+\sqrt{48}+.....$ is:

• A. $55$
• B. $66$
• C. $45\sqrt{3}$
• D. $55\sqrt{3}$

Answer 1

The correct option is D $55\sqrt{3}$

$\sqrt{3}+\sqrt{12}+\sqrt{27}+\sqrt{48}+\cdots +\sqrt{300}=\sqrt{3}+2\sqrt{3}+3\sqrt{3}+4\sqrt{3}+\cdots +10\sqrt{3}=\sqrt{3}(1+2+3+4+\cdots +10)$

$=\sqrt{3}(\frac{10\times 11}{2})=55\sqrt{3}$