Sequences & Series
Infinite Geometric Series
Grade 11
Question:
<p><strong>889.</strong> Let \(\alpha, \beta\) are the roots of the equation \(ax^2 + bx + c = 0\) where \(\beta = 4\alpha\) \((\alpha > 0)\). If \(3a = 2(c - b)\) and \(S = \displaystyle\sum_{r=0}^{\infty} \beta(\alpha^r)\), then find the value of \(3S\).</p>
Step-by-Step Solution
Key Concept: Use Vieta's formulas with the constraint β = 4α to express roots in terms of a single parameter, then apply the given algebraic condition 3a = 2(c - b) to find the ratio α/a, which determines the geometric series sum.
<p><strong>Step 1: Apply Vieta's formulas</strong></p><p>For roots α and β = 4α of ax² + bx + c = 0:</p><p>• Sum: α + 4α = -b/a → 5α = -b/a → b = -5aα</p><p>• Product: α · 4α = c/a → 4α² = c/a → c = 4aα²</p><p><strong>Step 2: Use the condition 3a = 2(c - b)</strong></p><p>Substitute c = 4aα² and b = -5aα:</p><p>3a = 2(4aα² - (-5aα))</p><p>3a = 2(4aα² + 5aα)</p><p>3a = 2a(4α² + 5α)</p><p>3 = 8α² + 10α</p><p>8α² + 10α - 3 = 0</p><p><strong>Step 3: Solve for α</strong></p><p>Using quadratic formula: α = (-10 ± √(100 + 96))/16 = (-10 ± 14)/16</p><p>Since α > 0: α = 4/16 = 1/4</p><p><strong>Step 4: Calculate the geometric series</strong></p><p>S = ∑(r=0 to ∞) β·α^r = β/(1-α)</p><p>Since β = 4α = 4(1/4) = 1:</p><p>S = 1/(1 - 1/4) = 1/(3/4) = 4/3</p><p><strong>Step 5: Find 3S</strong></p><p>3S = 3 · (4/3) = 4</p><p>∴ Answer: <strong>3S = 4</strong></p>
Correct Answer: 3