<p><strong>29.</strong> If \(a^2 + b^2\), \(ab + bc\), and \(b^2 + c^2\) are in G.P., then \(a, b, c\) are in</p>
Step-by-Step Solution
Key Concept: If three terms are in G.P., then (middle term)² = (first term)(third term). Apply this condition and factor to find the relationship between a, b, c that must hold for the constraint to be satisfied.
<p><strong>Step 1:</strong> For G.P., the middle term squared equals the product of extremes:</p><p>(ab + bc)² = (a² + b²)(b² + c²)</p><p><strong>Step 2:</strong> Expand left side: a²b² + 2ab²c + b²c²</p><p><strong>Step 3:</strong> Expand right side: a²b² + a²c² + b⁴ + b²c²</p><p><strong>Step 4:</strong> Equating and simplifying:</p><p>a²b² + 2ab²c + b²c² = a²b² + a²c² + b⁴ + b²c²</p><p>2ab²c = a²c² + b⁴</p><p><strong>Step 5:</strong> Divide by b²c (assuming b, c ≠ 0):</p><p>2ab = (a²c²/b²) + b²</p><p><strong>Step 6:</strong> Rearrange: a²c²/b² - 2ab + b² = 0, which is (ac/b - b)² = 0</p><p>Therefore: ac/b = b, or <strong>b² = ac</strong></p><p><strong>Step 7:</strong> The condition b² = ac is the definition of a, b, c being in G.P. (with common ratio b/a = c/b)</p><p>∴ Answer: B (a, b, c are in G.P.)</p>
Correct Answer: B