Let $A_{1}, A_{2}, A_{3}, \ldots$ be squares such that for each $n \geq 1$, the length of the side o

Question

Let $A_{1}, A_{2}, A_{3}, \ldots$ 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 $12 \mathrm{~cm}$, then the smallest value of $n$ for which area of $A_{n}$ is less than one, is

Professor PC Thomas Classes · Answer

(9): Given, length of side of $A_{1}=12$ Length of diagonal of $A_{2}=12$\ $ herefore \quad$ Length of side of $A_{2}=\frac{12}{\sqrt{2}}$\ Length of side of $A_{3}=\frac{12}{(\sqrt{2})^{2}}$\ Length of side of $A_{n}=\frac{12}{(\sqrt{2})^{n-1}}$\ $ herefore \quad$ Area of $A_{n}=\left(\frac{12}{(\sqrt{2})^{n-1}} ight)^{2}<1$\ $\Rightarrow \frac{144}{2^{(n-1)}}<1 \Rightarrow 2^{(n-1)}>144$\ $\Rightarrow n-1>7 \Rightarrow n>8$\ Since, $n$ is a natural number.\ $ herefore \quad$ Least value of $n$ is 9 .\