Explanation

(c)
Given, $x_{n} = x_{n + 1}\sqrt{2}$
$\therefore\ x_{1} = x_{2}\sqrt{2},\ x_{2} = x_{3}\sqrt{2,}\ldots,x_{n} = x_{n + 1}\sqrt{2}$
On multiplying $x_{1} = x_{n + 1}\left( \sqrt{2} \right)^{n}$
$\Rightarrow x_{n + 1} = x/\left( \sqrt{2} \right)^{n}$
Hence, $x_{n} = \frac{x_{1}}{\left( \sqrt{2} \right)^{n - 1}}$
Area of $S_{n} = x_{n}^{2} = \frac{x_{1}^{2}}{2^{n - 1}} < 1\ \Rightarrow 2^{n - 1} > x_{1}^{2}\ \ \ (\because x_{1} = 10)$
$\therefore\ \ 2^{n - 1} > 100$
But $2^{7} > 100,\ 2^{8} > 100$ etc.
$\therefore\ n - 1 = 7,8,\ 9,\ldots\ \Rightarrow n = 8,\ 9,\ 10,\ldots$