The sum of the geometric series $\sqrt{3}+\sqrt{12}+\sqrt{48}+\ldots$. up to 10 terms is\

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(a) : Let $S_{10}$ denotes the sum up to 10 terms of the given series. Then,\ $S_{10}=\sqrt{3}+\sqrt{12}+\sqrt{48}+\ldots \ldots \ldots$. up to 10 terms\ $=\sqrt{3}+2 \sqrt{3}+4 \sqrt{3}+\ldots \ldots \ldots 512 \sqrt{3}$\ Here, first term $a=\sqrt{3}$, common ratio, $r=2$, and number of terms, $n=10$