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 the sum of an infinite geometric series whose first three terms are \(a\), \(a^2\) and \(b\) respectively, is:</p>
<p>\(\dfrac{-1}{2}\)</p>
<p>\(\dfrac{-3}{2}\)</p>
<p>\(\dfrac{-1}{3}\)</p>
<p>none of these</p>
Step-by-Step Solution
Key Concept: In an arithmetic series, the common difference is constant, so b - a = a² - b. Use this to find the relationship between a and b, then verify the geometric series condition r = a²/a = b/a² before applying the infinite sum formula.
<p><strong>Step 1:</strong> Since a, b, a² form an AP with common difference d:</p><p>b - a = a² - b</p><p>2b = a + a²</p><p>b = (a + a²)/2</p><p><strong>Step 2:</strong> For a, a², b to form a GP, the common ratio must be constant:</p><p>a²/a = b/a²</p><p>a = b/a²</p><p>a³ = b</p><p><strong>Step 3:</strong> Substitute b = a³ into 2b = a + a²:</p><p>2a³ = a + a²</p><p>2a³ - a² - a = 0</p><p>a(2a² - a - 1) = 0</p><p>a(2a + 1)(a - 1) = 0</p><p>Since a ≠ 0: a = -1/2 or a = 1</p><p>Since a is negative: <strong>a = -1/2</strong></p><p><strong>Step 4:</strong> Then b = a³ = (-1/2)³ = -1/8</p><p>Common ratio of GP: r = a²/a = (1/4)/(-1/2) = -1/2</p><p>Since |r| = 1/2 < 1, the series converges.</p><p><strong>Step 5:</strong> Sum of infinite GP = a/(1 - r) = (-1/2)/(1 - (-1/2)) = (-1/2)/(3/2) = <strong>-1/3</strong></p><p>∴ Answer: B</p>
Correct Answer: B