Question:

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 10 cm, then for which of the following values of n is the area of $S_n$ less than 1 sq cm?

Updated On: Jun 14, 2022
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The Correct Option is D

Solution and Explanation

Let $a_n$ denotes the length of side of the square $S_n$.
We are given, $a_n$ = length of diagonal of $ S_{n+1}.$
$\Rightarrow \hspace20mm a_n = \sqrt 2\, a_{n+1} $
$\Rightarrow \hspace20mm a_{n+1} = \frac{a_n}{\sqrt 2}$
This shows that $ a_1 ,a_2, a_3,... $ form a GP with common ratio $1/ \sqrt 2.$
Therefore, $ a_n = a_1 \bigg(\frac{1}{\sqrt 2}\bigg)^{n-1} $
$\Rightarrow \hspace10mm a_n = 10 \bigg(\frac{1}{\sqrt 2}\bigg)^{n-1} \hspace10mm [\because a_1 = 10, given] $
$\Rightarrow \hspace10mm a^2_n = 100 \bigg(\frac{1}{\sqrt 2}\bigg)^{2(n-1)} $
$\Rightarrow \hspace10mm \frac{100}{2^{n-1}} \le 1 \hspace20mm [\because a^2_n \le 1, given]$
$\Rightarrow \hspace10mm 100 \le 2^{n-1}$
This is possible for $n \ge 8$.
Hence, (b), (c), (d) are the correct answers.
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