Sequences & Series
AP and GP
Grade 11
Question:
<p>If <i>a</i>, <i>b</i>, <i>c</i> are in AP and <i>a</i><sup>2</sup>, <i>b</i><sup>2</sup>, <i>c</i><sup>2</sup> are in GP such that <i>a</i> < <i>b</i> < <i>c</i> and \(a + b + c = \dfrac{3}{4}\), then the value of <i>a</i> is</p>
<p>\(\dfrac{1}{4} - \dfrac{1}{4\sqrt{2}}\)</p>
<p>\(\dfrac{1}{4} - \dfrac{1}{3\sqrt{2}}\)</p>
<p>\(\dfrac{1}{4} - \dfrac{1}{2\sqrt{2}}\)</p>
<p>\(\dfrac{1}{4} - \dfrac{1}{\sqrt{2}}\)</p>
Step-by-Step Solution
Key Concept: Since a, b, c are in AP, we have b = (a+c)/2. Combined with the GP condition b² = a²·c² and the sum constraint, we can express everything in terms of the common difference to create a solvable system.
<p><strong>Step 1:</strong> Since a, b, c are in AP: b = (a+c)/2, so a + c = 2b</p><p><strong>Step 2:</strong> From a + b + c = 3/4 and a + c = 2b: 2b + b = 3/4 ⟹ b = 1/4</p><p><strong>Step 3:</strong> Since a², b², c² are in GP: b⁴ = a²c² ⟹ (1/4)⁴ = a²c² ⟹ a²c² = 1/256</p><p><strong>Step 4:</strong> We have c = (3/4 - a) and a(3/4 - a) = 1/256 (since ac = 1/16, taking positive root for a < b < c)</p><p><strong>Step 5:</strong> Expanding: 3a/4 - a² = 1/16 ⟹ 12a - 4a² = 1 ⟹ 4a² - 12a + 1 = 0</p><p><strong>Step 6:</strong> Using quadratic formula: a = (12 ± √(144-16))/8 = (12 ± √128)/8 = (12 ± 8√2)/8 = (3 ± 2√2)/2</p><p><strong>Step 7:</strong> Since a < b = 1/4, we need a = (3 - 2√2)/2 ≈ 0.086 (the other root gives a > 1/4)</p><p>∴ Answer: C</p>
Correct Answer: C