Question

Let A₁, A₂, A₃, … be squares such that for each n ≥ 1, the length of the side of $A_n$ equals the length of diagonal of $A_{n+1}$. If the length of $A_1$ is 12 cm, then the smallest value of n for which area of $A_n$ is less than one, is __________.

Hint:

If the side of a square is the diagonal of the next, the areas form a geometric progression with a common ratio of $1/2$.

Answer 1

Correct Answer: 9

Step 1: Side of $A_n = s_n$. Diagonal of $A_{n+1} = \sqrt{2} s_{n+1}$.
Step 2: Given $s_n = \sqrt{2} s_{n+1} \Rightarrow s_{n+1} = \frac{s_n}{\sqrt{2}}$.
Step 3: This is a G.P. with $s_1 = 12$ and $r = 1/\sqrt{2}$.
Step 4: Area $a_n = s_n^2$. $a_1 = 144$. $a_{n+1} = s_{n+1}^2 = \frac{s_n^2}{2} = \frac{a_n}{2}$.
Step 5: $a_n = 144 \cdot (\frac{1}{2})^{n-1}$.
Step 6: Set $a_n<1 \Rightarrow \frac{144}{2^{n-1}}<1 \Rightarrow 2^{n-1}>144$.
Step 7: $2^7 = 128, 2^8 = 256$. So $n-1 = 8 \Rightarrow n=9$.