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 $12$ cm, then the smallest value of $n$ for which area of $A_n$ is less than one, is

Answer 1

$\therefore$ side lengths are in G.P.
$T_n=\frac{12}{(\sqrt{2}{)}^{n-1}}$
$\therefore$ Area $=\frac{144}{2^{n-1}}<1$ $\Rightarrow 2^{n-1}>144$
Smallest $n=9$