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 $10$ cm, then the least value of $n$ for which the area of $S_n$ less that 1 sq cm

• A. $7$
• B. $8$
• C. $9$
• D. $10$

Answer 1

The correct option is B $8$

$S_1=10$ cm, $\text{Area}\left(S_n\right)<100{\text{cm}}^2$
Length of $S_n=$ length of diagonal of $S_{n+1}$
As we know, length of $S_n=\sqrt{2}$ length of side of $S_{n+1}$
$\therefore \frac{\text{length of side}{S}_{n+1}}{\text{length of side}{S}_n}=\frac{\text{length of side}{S}_n}{\text{length of side}{S}_{n-1}}=.....=\frac{\text{length of side}{S}_2}{\text{length of side}{S}_1}=\frac{1}{\sqrt{2}}$
Therefore, $S_1,S_2,S_3......,S_{n+1}$ are in GP
$\therefore a=10,r=\frac{1}{\sqrt{2}}$
Sides of $S_n=10{\left(\frac{1}{\sqrt{2}}\right)}^{n-1}=\frac{10}{2^{(n-1)/2}}$
Area $={\left(\frac{10}{2^{(n-1)/2}}\right)}^2<1$