Question

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

• A. $7$
• B. $8$
• C. $9$
• D. $10$

Answer 1

The correct option is B $8$

Given, length of a side of $S_n$.
$=$ Length of a diagonal of $S_{n+1}$
$\Rightarrow$ Length of a side of $S_n$
$=\sqrt{2}$ (Length of a side of $S_{n+1})$
$\Rightarrow \frac{\text{Length of a side of}{S}_{n+1}}{\text{Length of a side of}{S}_n}=\frac{1}{\sqrt{2}}$ for all $n\ge 1$
So, the side of $S_1,S_2,....,S_n$ from a $GP$, with common ration is $\frac{1}{\sqrt{2}}$ and first term $10$.
$\therefore$ Side of $S_n=10{\left(\frac{1}{\sqrt{2}}\right)}^{n-1}=\frac{10}{2^{\frac{(n-1)}{2}}}$
Since, area of $S_n<1$ (given)
$\Rightarrow \frac{100}{2^{n-1}}<1$
$\Rightarrow 2^{n-1}>100$
$\Rightarrow n-1\ge 7\Rightarrow n\ge 8$.