<p>Let \(S_1, S_2, \ldots\) be squares such that for each \(n \geq 1\), 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?</p>
Step-by-Step Solution
Key Concept: The side length forms a geometric sequence with common ratio 1/√2, since each side equals the diagonal of the next square (diagonal = side × √2). Track the area decrease using the quadratic relationship: if side of S_n is a_n, then area = a_n², and a_n = 10/(√2)^(n-1).
<p><strong>Step 1:</strong> Set up the relationship. If side of S_n is a_n, then diagonal of S_n is a_n√2. Given: a_n = a_{n+1}√2, so a_{n+1} = a_n/√2.</p><p><strong>Step 2:</strong> This is a geometric sequence with first term a_1 = 10 and common ratio r = 1/√2. Thus a_n = 10 · (1/√2)^(n-1) = 10 · 2^{-(n-1)/2}.</p><p><strong>Step 3:</strong> Area of S_n is A_n = a_n² = 100 · 2^{-(n-1)} = 100/2^(n-1).</p><p><strong>Step 4:</strong> Find when A_n < 1: 100/2^(n-1) < 1 ⟹ 2^(n-1) > 100 ⟹ (n-1)log2 > log100 ⟹ n-1 > log₂(100) ≈ 6.64 ⟹ n > 7.64.</p><p><strong>Step 5:</strong> Therefore A_n < 1 when n ≥ 8. The values satisfying this condition are n = 8, 9, 10, ... (and likely bounded by the answer choices given as B, C, D).</p><p>∴ Answer: B, C, D</p>
Correct Answer: B,C,D