Question

Let $S_1,S_2,S_3,...$be squares such that for each $n\ge 1$, the lenght 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 $10$cm and the area of $S_n$$<1$sq.cm.The the value of $n$ is

• A. $n>7$
• B. $n\ge 8$
• C. $n<7$
• D. $n\le 8$

Answer 1

Answer: (A), (B)

The correct options are
A $n>7$
B $n\ge 8$

We have,length of a side of $S_n=$length of the diagonal of $S_{n+1}$

$\Rightarrow$length of a side of $S_n=\sqrt{2}\times$length of a side of $S_{n+1}$

$\Rightarrow \frac{lengthofasideof{S}_{n+1}}{lengthofasideof{S}_n}=\frac{1}{\sqrt{2}}$ for all $n\ge 1$

$\Rightarrow$sides of $S_1,s_2,S_3,...$ form a G.P with common ratio $\frac{1}{\sqrt{2}}$ and the first term $10$

$\therefore$Side of $S_n=10{\left(\frac{1}{\sqrt{2}}\right)}^{n-1}=\frac{10}{2^{\frac{n-1}{2}}}$

$\Rightarrow$Area of $S_n={\left(side\right)}^2={\left(\frac{10}{2^{\frac{n-1}{2}}}\right)}^2=\frac{100}{2^{n-1}}$

Now,given area of $S_n<1$

$\Rightarrow \frac{100}{2^{n-1}}<1$

$\Rightarrow 2^{n-1}>100>2^6$

$\Rightarrow n-1>6$

$\Rightarrow n>7$ or $n\ge 8$