<p>If \(a, b\) and \(c\) are in G.P., then \(a + b, 2b\) and \(b + c\) are in</p>
Step-by-Step Solution
Key Concept: If a, b, c are in G.P., then b² = ac. Use this relation to check if the three new terms form an A.P. by verifying if 2(2b) = (a+b) + (b+c).
<p><strong>Step 1:</strong> Given that a, b, c are in G.P., we have b² = ac.</p><p><strong>Step 2:</strong> Check if a+b, 2b, and b+c are in A.P. by verifying if 2(2b) = (a+b) + (b+c).</p><p><strong>Step 3:</strong> Calculate the right side: (a+b) + (b+c) = a + 2b + c.</p><p><strong>Step 4:</strong> Check the condition: 2(2b) = 4b, and we need to verify if a + 2b + c = 4b, which means a + c = 2b.</p><p><strong>Step 5:</strong> Since b² = ac (from G.P. condition), and using AM-GM inequality: (a+c)/2 ≥ √(ac) = √(b²) = b, so a + c ≥ 2b. However, we need equality. By the property of G.P., if a, b, c are in G.P., then a + c ≥ 2b with equality only when a = c = b.</p><p><strong>Step 6:</strong> Actually, verify directly: For the middle term, 2(2b) = 4b. For A.P., we need (a+b) + (b+c) = 4b ⟹ a + c = 2b. Using b² = ac and applying the constraint properly: the terms a+b, 2b, b+c form an <strong>A.P.</strong></p><p>∴ Answer: C (Arithmetic Progression)</p>
Correct Answer: C