A) 7
B) 8
C) 9
D) 10
Correct Answer: C
Solution :
Given \[{{x}_{n}}={{x}_{n+1}}\sqrt{2}\] \[\therefore \]\[{{x}_{1}}={{x}_{2}}\sqrt{2},\ {{x}_{2}}={{x}_{3}}\sqrt{2},\ {{x}_{n}}={{x}_{n+1}}\sqrt{2}\] On multiplying \[{{x}_{1}}={{x}_{n+1}}{{(\sqrt{2})}^{n}}\]\[\Rightarrow \]\[{{x}_{n+1}}=\frac{{{x}_{1}}}{{{(\sqrt{2})}^{n}}}\] Hence \[{{x}_{n}}=\frac{{{x}_{1}}}{{{(\sqrt{2})}^{n-1}}}\] Area of \[{{S}_{n}}=x_{n}^{2}=\frac{x_{1}^{2}}{{{2}^{n-1}}}<1\]\[\Rightarrow \]\[{{2}^{n-1}}>x_{1}^{2}\ \ \ \ \ \ ({{x}_{1}}=10)\] \[\therefore \]\[{{2}^{n-1}}>100\] But \[{{2}^{7}}>100,\ {{2}^{8}}>100,\ \]etc. \[\therefore \]\[n-1=7,\ 8,\ 9.....\]\[\Rightarrow \]\[n=8,\ 9,\ 10.....\]
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