Question

Let $A _{1}, A _{2}, A _{3}, \ldots \ldots . .$ be squares such that for each $n \geq 1,$ the length of the side of $A _{ n }$ equals the length of diagonal of $A _{ n +1}$. If the length of $A _{1}$ is $12 cm$, then the smallest value of $n$ for which area of $A _{ n }$ is less than one, is_____

Answer 1

Let $a_{n}$ be the side length of $A_{n}$
So, $a_{n}=\sqrt{2} a_{n+1}, a_{1}=12$
$\Rightarrow a_{n}=12 \times\left(\frac{1}{\sqrt{2}}\right)^{n-1}$
Now, $\left(a_{n}\right)^{2}<1 \Rightarrow \frac{144}{2^{(n-1)}}<1$
$\Rightarrow 2^{(n-1)}>144$
$\Rightarrow n -1 \geq 8$
$\Rightarrow n \geq 9$