Sequences & Series
AP-GP
Grade 11
Question:
<p>For any three positive real numbers \(a\), \(b\) and \(c\), \(9(25a^2 + b^2) + 25(c^2 - 3ac) = 15b(3a + c)\). Then</p>
<p>\(a\), \(b\) and \(c\) are in G.P.</p>
<p>\(b\), \(c\) and \(a\) are in G.P.</p>
<p>\(b\), \(c\) and \(a\) are in A.P.</p>
<p>\(a\), \(b\) and \(c\) are in A.P.</p>
Step-by-Step Solution
Key Concept: Rearrange the constraint equation into a sum of squares form. This reveals that multiple terms must simultaneously equal zero, forcing specific relationships between a, b, and c.
<p><strong>Step 1:</strong> Expand and 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> Rearrange as a sum of squares:</p><p>225a² + 9b² + 25c² - 75ac - 45ab - 15bc = 0</p><p>= (15a)² + (3b)² + (5c)² - 2(15a)(3b)/2 - 2(15a)(5c)/2 - 2(3b)(5c)/2</p><p>= (15a - 3b)² + (5c)² - 2(15a - 3b)(5c) = 0</p><p>= (15a - 3b - 5c)² = 0</p><p><strong>Step 3:</strong> Since the sum of squares equals zero, each term must be zero:</p><p>15a - 3b - 5c = 0</p><p>∴ 15a = 3b + 5c, which gives us the relationship between a, b, and c</p><p>This means: <strong>b : c = 3 : 1</strong> when a is expressed in terms of b and c, or <strong>a : b : c = 1 : 3 : 5</strong> in simplest form.</p>
Correct Answer: D