Question

Let $S_1,S_2,S_3,...$ 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 $10$cm, then for which of the following values of $n$ is the area of $S_n$ less than $1$sq.cm?

• A. $7$
• B. $8$
• C. $9$
• D. $10$

Answer 1

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

The correct options are
B $8$
C $9$
D $10$

If $a$ be the side of the square, then diagonal $d=a\sqrt{2}$.
By hypothesis
$\Rightarrow a_{n+1}=\frac{a_n}{\sqrt{2}}=\frac{a_{n-1}}{{\left(\sqrt{2}\right)}^2}=\frac{a_{n-2}}{{\left(\sqrt{2}\right)}^3}=...=\frac{a_1}{{\left(\sqrt{2}\right)}^n}=$
$\therefore a_{n+1}=\frac{a_1}{{\left(\sqrt{2}\right)}^n}$
$\Rightarrow a_n=\frac{a_1}{{\left(\sqrt{2}\right)}^{n-1}}=\frac{10}{2^{\frac{(n-1)}{2}}}$
$\because$ Area of $S_n<1\Rightarrow a_n^2<1$
$\Rightarrow \frac{100}{2^{n-1}}<1$
$\Rightarrow 2^n>200>2^7$
$\Rightarrow 2^n>2^7$
$\Rightarrow n>7$ (Since bases are same we can equate the powers)
$\therefore n=8,9,10$