11th Class Mathematics Other Series Question Bank Critical Thinking

  • question_answer Let \[{{S}_{1}},\ {{S}_{2}},.......\]be squares such that for each \[n\ge 1\], the length of a side of \[{{S}_{n}}\] equals the length of a diagonal of \[{{S}_{n+1}}\]. If the length of a side of \[{{S}_{1}}\]is\[10cm\], then for which of the following values of \[n\] is the area of \[{{S}_{n}}\] less then \[1\ sq\ cm\] [IIT 1999]

    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. \[\therefore \]\[n-1=7,\ 8,\ 9.....\]\[\Rightarrow \]\[n=8,\ 9,\ 10.....\]

You need to login to perform this action.
You will be redirected in 3 sec spinner