The sum to 10 terms of the series $\sqrt{2} + \sqrt{6} + \sqrt{18} + \sqrt{54} + \dots$ is

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$243 (\sqrt{3} + 1)$

$\frac{121}{2} (\sqrt{3} + 1)$

$243 (\sqrt{3} - 1)$

$121 (\sqrt{6} + \sqrt{2})$

answer is A.

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Detailed Solution

$\begin{array}{l} a = \sqrt{2}, r = \sqrt{3} \\ S_{10} = \frac{a (r^{10} - 1)}{r - 1} = \frac{\sqrt{2} (3^{5} - 1)}{\sqrt{3} - 1} \\ = \frac{\sqrt{2} (242) (\sqrt{3} + 1)}{(\sqrt{3} - 1) (\sqrt{3} + 1)} = 121 (\sqrt{6} + \sqrt{2}) \end{array}$

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