Question:
Let $\mathrm{A}_{1}, \mathrm{~A}_{2}, \mathrm{~A}_{3}, \ldots \ldots . .$ 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 $\mathrm{A}_{1}$ is $12 \mathrm{~cm}$, then the smallest value of $\mathrm{n}$ for which area of $A_{n}$ is less than one, is____________.
Solution:
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, $\left(a_{n}\right)^{2}<1 \Rightarrow \frac{144}{2^{(n-1)}}<1$
$\Rightarrow 2^{(n-1)}>144$
$\Rightarrow n-1 \geq 8$
$\Rightarrow n \geq 9$