Sequences & Series
Arithmetic and Geometric Progressions
Grade 11
Question:
<p>The 1st, 2nd and 3rd terms of an arithmetic series are \(a\), \(b\) and \(a^2\), where \(a\) is negative. Then sum of an infinite geometric series whose first three terms are \(a\), \(a^2\) and \(b\) respectively, is:</p>
<p>(a) \(\dfrac{-1}{2}\)</p>
<p>(b) \(\dfrac{-3}{2}\)</p>
<p>(c) \(\dfrac{-1}{3}\)</p>
<p>(d) none of these</p>
Step-by-Step Solution
Key Concept: Use the arithmetic progression property (2nd term is average of 1st and 3rd) to find relationships between a and b, then verify the geometric series condition before applying the infinite sum formula.
<p><strong>Step 1: Use AP property for first series</strong></p><p>If a, b, a² are in AP, then the middle term equals the average:</p><p>2b = a + a²</p><p>∴ b = (a + a²)/2</p><p><strong>Step 2: Verify the second series is geometric</strong></p><p>The three terms are a, a², b. For a geometric series:</p><p>(a²)² = a·b</p><p>a⁴ = a·(a + a²)/2</p><p>a⁴ = (a² + a³)/2</p><p>2a⁴ = a² + a³</p><p>2a⁴ - a³ - a² = 0</p><p>a²(2a² - a - 1) = 0</p><p>a²(2a + 1)(a - 1) = 0</p><p>Since a ≠ 0 and a is negative: a = -1/2</p><p><strong>Step 3: Find common ratio and sum</strong></p><p>First term of GP: a = -1/2</p><p>Second term: a² = 1/4</p><p>Common ratio: r = (1/4)/(-1/2) = -1/2</p><p>Since |r| = 1/2 < 1, the infinite sum exists:</p><p>S = a/(1 - r) = (-1/2)/(1 - (-1/2)) = (-1/2)/(3/2) = -1/3</p><p>∴ Answer: A (which should be -1/3)</p>
Correct Answer: A