Question

Let $S_1,S_2,\dots$ be 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 $10\text{cm},$ then for which of the following values of $n$ is the area of $S_n$ less than $1\text{sq. cm}?$

• A. $7$
• B. $8$
• C. $9$
• D. $10$

Answer 1

Answer: (B), (C), (D)

$10$

We have, length of a side of $S_n$ is equal to the length of a diagonal of $S_{n+1}$.
Hence,
Length of a side of $S_n=\sqrt{2}$(Length of a side of $S_{n+1}$)
$\Rightarrow \frac{\text{Length of a side of}{S}_{n+1}}{\text{Length of side of}{S}_n}=\frac{1}{\sqrt{2}}$, for all $n\ge 1$
Hence, sides of $S_1,S_2,\dots ,S_n$ form a G.P with common ratio $\frac{1}{\sqrt{2}}$ and first term $10.$

$\therefore$ Side of $S_n=10{\left(\frac{1}{\sqrt{2}}\right)}^{n-1}=\frac{10}{2^{(n-1)/2}}$
$\Rightarrow$ Area of $S_n=(\text{side}{)}^2={\left(\frac{10}{2^{(n-1)/2}}\right)}^2=\frac{100}{2^{n-1}}$
Now, area of $S_n<1$
$\Rightarrow \frac{100}{2^{n-1}}<1$
$\Rightarrow 100<2^{n-1}$
$\Rightarrow n>7$