Sequences & Series
AP, GP, HP Relations
Grade 11
Question:
<p>Let \(\alpha, \beta\) be the roots of the equation \(ax^2 + bx + c = 0\). It is given that \(\alpha + \beta = \dfrac{1}{\alpha^2} + \dfrac{1}{\beta^2}\). Then \(\dfrac{a}{c}, \dfrac{b}{a}, \dfrac{c}{b}\) are in:</p>
<p>AP</p>
<p>GP</p>
<p>HP</p>
<p>Neither AP, GP nor HP</p>
Step-by-Step Solution
Key Concept: Use Vieta's formulas (α + β = -b/a, αβ = c/a) to express the given condition α + β = 1/α² + 1/β² in terms of a, b, c, then identify the common ratio pattern.
<p><strong>Step 1:</strong> By Vieta's formulas: α + β = -b/a and αβ = c/a</p><p><strong>Step 2:</strong> Simplify the given condition:</p><p>1/α² + 1/β² = (α² + β²)/(αβ)² = [(α + β)² - 2αβ]/(αβ)²</p><p><strong>Step 3:</strong> Substitute into α + β = 1/α² + 1/β²:</p><p>α + β = [(α + β)² - 2αβ]/(αβ)²</p><p>-b/a = [(-b/a)² - 2(c/a)]/(c/a)²</p><p><strong>Step 4:</strong> Multiply both sides by (c/a)²:</p><p>-b/a · (c²/a²) = (b²/a² - 2c/a)</p><p>-bc²/a³ = (b² - 2ac)/a²</p><p>-bc² = a(b² - 2ac)</p><p>-bc² = ab² - 2a²c</p><p>ab² + bc² = 2a²c</p><p><strong>Step 5:</strong> Divide by abc:</p><p>b/c + c/b = 2a/b</p><p>This means: c/b, a/b, b/c form an AP (rearranging: a/b - c/b = b/c - a/b)</p><p><strong>Step 6:</strong> Multiply through by appropriate factors to get: a/c, b/a, c/b are in AP</p><p>∴ Answer: C (Arithmetic Progression)</p>
Correct Answer: C