Question

Let $S_1,S_2\dots$be the squares such that for each $n\ge 1$, the length of a side of $S_n$ equals the length of a diagonal of $S_{n+1}$. If the length of a side of $S_1$ is $10cm$, then for which of the following values of $n$ is the area of $S$ less than $1sqcm$?

• A. $7$
• B. $8$
• C. $9$
• D. $10$

Answer 1

The correct option is B

$8$

Explanation for the correct option:

Finding the values of $n$:

Considering $a$ be the length of the sides of the square.

Therefore diagonal of the square be $d=a\sqrt{2}$

Given that $a_n=$ diagonal of $S_{n+1}$

$$\begin{gathered} \Rightarrow a_n=\sqrt{2}{a}_{n+1} \\ \Rightarrow a_{n+1}=\frac{a_n}{\sqrt{2}}...\left(i\right) \end{gathered}$$

It shows a GP with a common ratio $r=\frac{1}{\sqrt{2}}$

Since given that $a_1=10$

$$\begin{gathered} \Rightarrow a_n=a_1{\left(\frac{1}{\sqrt{2}}\right)}^{n-1} \\ \Rightarrow a_n=10{\left(\frac{1}{\sqrt{2}}\right)}^{n-1} \\ \Rightarrow {\left(a_n\right)}^2=100{\left(\frac{1}{2^{{\displaystyle \frac{1}{2}}}}\right)}^{2\left(n-1\right)}\left[\text{squaring both sides}\right] \\ \Rightarrow \frac{100}{2^{n-1}}\le 1\left[\because {\left(a_n\right)}^2\le 1\text{is given}\right] \end{gathered}$$

It is possible if $n\ge 8$

.Hence, option (B) is the correct answer.