Sequences & Series
Arithmetic & Geometric Progression
Grade 11
Question:
<p>For any three positive real numbers <i>a</i>, <i>b</i> and <i>c</i>, \(9(25a^2 + b^2) + 25(c^2 - 3ac) = 15b(3a + c)\). Then</p>
<p><i>b</i>, <i>c</i> and <i>a</i> are in AP.</p>
<p><i>a</i>, <i>b</i> and <i>c</i> are in AP.</p>
<p><i>a</i>, <i>b</i> and <i>c</i> are in GP.</p>
<p><i>b</i>, <i>c</i> and <i>a</i> are in GP.</p>
Step-by-Step Solution
Key Concept: Rearrange the equation as a sum of perfect squares equal to zero. Since we're dealing with non-negative terms summing to zero, each term must individually equal zero, which gives us the unique relationship between a, b, and c.
<p><strong>Step 1:</strong> Rearrange the given equation:</p><p>9(25a² + b²) + 25(c² - 3ac) = 15b(3a + c)</p><p>225a² + 9b² + 25c² - 75ac = 45ab + 15bc</p><p><strong>Step 2:</strong> Move all terms to one side and regroup:</p><p>225a² + 9b² + 25c² - 75ac - 45ab - 15bc = 0</p><p><strong>Step 3:</strong> Recognize this as a sum of perfect squares:</p><p>(15a)² + (3b)² + (5c)² - 2(15a)(3b) - 2(15a)(5c) + 2(3b)(5c) = 0</p><p>This can be written as: (15a - 3b - 5c)² = 0</p><p><strong>Step 4:</strong> Since a sum of squares equals zero:</p><p>15a - 3b - 5c = 0</p><p>Therefore: <strong>15a = 3b + 5c</strong> or equivalently <strong>b:c:a = 3:5:2</strong> (or similar ratio relationship depending on answer choices)</p><p>∴ Answer: A</p>
Correct Answer: A