Question

Let $A_1,A_2,A_3,\dots .$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

Step 1: Find the length of the square:

Let we assume the length of the side of $A_n=l_n$

For each $n\ge 1$, the length of the side of $A_n$ equals the length of diagonal of $A_{n+1}$.

By using the Pythagoras theorem, the length of the diagonal of the square is calculated as:

$$\begin{gathered} l_n=\sqrt{{\left(l_{n+1}\right)}^2+{\left(l_{n+1}\right)}^2} \\ =\sqrt{2}{l}_{n+1} \end{gathered}$$

$$\begin{gathered} \therefore \frac{l_{n+1}}{l_n}\text{or}r=\frac{l_{n+1}}{\sqrt{2}{l}_{n+1}}\left[[\text{where},r=\text{common ratio}\right] \\ =\frac{1}{\sqrt{2}} \end{gathered}$$

Therefore, the sides of the length are in Geometric progression.

General term of G.P. is $T_n=a{r}^{n-1},\text{where}a=\text{first term and}r=\text{common ratio}$

The length of $A_1$ is $12cm$ given and $r=\frac{1}{\sqrt{2}}$

$\therefore T_n=\left(12\right){\left(\frac{1}{\sqrt{2}}\right)}^{n-1}$

Step 2: Find the smallest value of $n$:

Area of the square = Square of the length of the square.

Therefore, Area is given as:

$$\begin{gathered} ar\left(Square\right)={\left(\left(12\right){\left(\frac{1}{\sqrt{2}}\right)}^{n-1}\right)}^2 \\ =\frac{144}{2^{n-1}} \end{gathered}$$

The smallest value of $n$, for which Area is less than one would be given as:

$$\begin{gathered} \frac{144}{2^{n-1}}<1 \\ \Rightarrow 144<2^{n-1} \\ \Rightarrow 144<256 \\ \Rightarrow 144<2^8 \\ For{2}^8=2^{n-1} \end{gathered}$$

Equating the powers, we have

$$\begin{gathered} \Rightarrow n-1=8 \\ \Rightarrow n=9 \end{gathered}$$

Hence the smallest value of $n=9$.