$\begin{array}{1 1} 11 \\ 21 \\ 10 \\ 20 \end{array}$

Answer : (c) 10

Explanation : $\sqrt2,2,2\sqrt2$ is a GP with $a\sqrt2\quad\;r=\sqrt2$

$S_{n}=\large\frac{\sqrt2((\sqrt2)^n-1)}{\sqrt2-1}=62+31\sqrt2$

$\sqrt2 \;(\sqrt2^n-1)=(62+31\sqrt2)(\sqrt2-1)$

$\sqrt2 \;(\sqrt2^n-1)=62\;\sqrt2-62+62-31\sqrt2$

$\sqrt2 \;(\sqrt2^n-1)=31\sqrt2$

$ \sqrt2^n-1=31$

$ \sqrt2^n=32$

$n=10.$

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