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Let $S_{1}, S_{2}, \dots$ be squares such that for each $n \geq 1$ , 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 S₁, is 10 cm, then for which of the following values of n, the area of Sn, less than 1 sq cm ?

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None of these

answer is C.

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Detailed Solution

Given, $x_{n} = x_{n + 1} \sqrt{2}$

$∴ x_{1} = x_{2} \sqrt{2}, x_{2} = x_{3} \sqrt{2}, \dots, x_{n} = x_{n + 1} \sqrt{2}$

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 $S_{n} = x_{n}^{2} = \frac{x_{1}^{2}}{2^{n - 1}} < 1$

$\Rightarrow 2^{n - 1} > x_{1}^{2} (∵ x_{1} = 10)$

$∴ 2^{n - 1} > 100$

But, $2^{7} > 100, 2^{8} > 100$ etc.

$∴ n - 1 = 7, 8, 9, \dots \Rightarrow n = 8, 9, 10, \dots$

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