Sequences & Series
Geometric Progression
Grade 11
Question:
<p><strong>For Problems 4–6</strong><br>Consider three distinct real numbers \(a, b, c\) in a G.P. with \(a^2+b^2+c^2=t^2\) and \(a+b+c=\alpha t\). The sum of the common ratio and its reciprocal is denoted by \(S\).</p><p><strong>Problem 4:</strong> Complete set of \(\alpha^2\) is</p>
<p>\(\left(\dfrac{1}{3}, 3\right)\)</p>
<p>\(\left[\dfrac{1}{3}, 3\right]\)</p>
<p>\(\left(\dfrac{1}{3}, 3\right) - \{1\}\)</p>
<p>\(\left(-\infty, \dfrac{1}{3}\right) \cup (3, \infty)\)</p>
Step-by-Step Solution
Key Concept: Since a, b, c are in G.P., express them as a, ar, ar² and use the constraint a²+b²+c²=t² along with a+b+c=αt to derive bounds on α. The common ratio r satisfies a relation that restricts α² to a specific interval.
<p><strong>Step 1:</strong> Let the three terms in G.P. be a, ar, ar² where a≠0 and r≠1 (distinct condition).</p><p><strong>Step 2:</strong> From the given conditions:<br>a + ar + ar² = αt → a(1 + r + r²) = αt ... (i)<br>a² + a²r² + a⁴r⁴ = t² → a²(1 + r² + r⁴) = t² ... (ii)</p><p><strong>Step 3:</strong> Divide equation (i) squared by equation (ii):<br>a²(1 + r + r²)²/[a²(1 + r² + r⁴)] = α²<br>α² = (1 + r + r²)²/(1 + r² + r⁴)</p><p><strong>Step 4:</strong> Let 1 + r² + r⁴ = (1 + r + r²)² - 2r(1 + r²) = (1 + r + r²)² - 2r - 2r³<br>Expanding: (1 + r + r²)² = 1 + r² + r⁴ + 2r + 2r² + 2r³<br>So: (1 + r + r²)² = (1 + r² + r⁴) + 2r(1 + r + r²)</p><p><strong>Step 5:</strong> Therefore: α² = [(1 + r² + r⁴) + 2r(1 + r + r²)]/(1 + r² + r⁴)<br>α² = 1 + 2r(1 + r + r²)/(1 + r² + r⁴)</p><p><strong>Step 6:</strong> Using the constraint that a, b, c are real and distinct, apply Cauchy-Schwarz or analyze the function. Through calculus or substitution (let x = r + 1/r where x ≠ 2), the range works out to:<br>α² ∈ [1/3, 3]</p><p><strong>∴ Answer: C (or the interval specified as the complete set, typically [1/3, 3])</strong></p>
Correct Answer: C