Sequences & Series
Harmonic Progression
Grade 11
Question:
<p>If the sum of the roots of the quadratic equation \(ax^2 + bx + c = 0\) is equal to the sum of the squares of their reciprocals, then \(\dfrac{a}{c}\), \(\dfrac{b}{a}\) and \(\dfrac{c}{b}\) are in</p>
<p>arithmetic progression.</p>
<p>geometric progression.</p>
<p>harmonic progression.</p>
<p>arithmetic-geometric-progression.</p>
Step-by-Step Solution
Key Concept: Use Vieta's formulas to express the sum of roots and sum of squares of reciprocals in terms of coefficients, then establish the relationship between the three given expressions.
<p><strong>Step 1:</strong> Let the roots be α and β. By Vieta's formulas: α + β = -b/a and αβ = c/a</p><p><strong>Step 2:</strong> Sum of squares of reciprocals: (1/α)² + (1/β)² = (α² + β²)/(αβ)² = [(α+β)² - 2αβ]/(αβ)²</p><p><strong>Step 3:</strong> Substituting: [(−b/a)² − 2(c/a)]/(c/a)² = [b²/a² − 2c/a]/c²/a² = (b² − 2ac)/c²</p><p><strong>Step 4:</strong> Given condition: α + β = (1/α)² + (1/β)²</p><p>−b/a = (b² − 2ac)/c²</p><p><strong>Step 5:</strong> Cross-multiplying: −bc² = a(b² − 2ac) = ab² − 2a²c</p><p>−bc² = ab² − 2a²c</p><p>2a²c − bc² = ab²</p><p><strong>Step 6:</strong> Dividing by abc: 2a/b − c/a = b/c</p><p>Rearranging: a/c, b/a, c/b form an AP when 2(b/a) = (a/c) + (c/b)</p><p>This simplifies to show: a/c, b/a, c/b are in <strong>Arithmetic Progression (AP)</strong></p><p>∴ Answer: C</p>
Correct Answer: C