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Let A1, A2, A3......be squares such that for each n\geq 1, the length of the side An equals the length of diagonal of  An+1. If the length of A1 is 12 cm, then the smallest value of n for which area of An is less one, is _______
Option: 1 5
Option: 2 7
Option: 3 9
Option: 4 11

Answers (1)

best_answer

 

\begin{aligned} &\text { Let } a_{n} \text { be the side length of } A_{n}\\ &\text { So, } a_{n}=\sqrt{2} a_{n+1}, a_{1}=12\\ &\Rightarrow a_{n}=12 \times\left(\frac{1}{\sqrt{2}}\right)^{n-1} \end{aligned}

\begin{aligned} \text {Now, }\left(a_{n}\right)^{2}<1 &\Rightarrow \frac{144}{2^{(n-1)}}<1 \\ \Rightarrow 2^{(n-1)}>144 &\\ \Rightarrow n-1 \geq 8 &\\ \Rightarrow n \geq 9 \end{array}

 

Posted by

Suraj Bhandari

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