Question

Let $A_1,A_2,A_3,\dots \dots ..$ be squares such that for each $n\ge 1,$ the length of the side of $A_n$ equals the length of diagonal of $A_{n+1}$. If the length of $A_1$ is $12cm$, then the smallest value of $n$ for which area of $A_n$ is less than one, is_____

Answer 1

Answer: 9

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, ${\left(a_n\right)}^2<1\Rightarrow \frac{144}{2^{(n-1)}}<1$
$\Rightarrow 2^{(n-1)}>144$
$\Rightarrow n-1\ge 8$
$\Rightarrow n\ge 9$