<p>\(a, b, c, d \in R^+\) such that \(a, b\) and \(c\) are in A.P. and \(b, c\) and \(d\) are in H.P., then</p>
Step-by-Step Solution
Key Concept: Use the A.P. condition to express b and c in terms of a, then convert the H.P. condition to an A.P. condition for reciprocals to establish relationships between all four terms.
<p><strong>Step 1:</strong> From a, b, c in A.P.: 2b = a + c, so c = 2b - a</p><p><strong>Step 2:</strong> From b, c, d in H.P.: 1/b, 1/c, 1/d are in A.P., so 2/c = 1/b + 1/d</p><p><strong>Step 3:</strong> Substitute c = 2b - a into the H.P. condition: 2/(2b - a) = 1/b + 1/d</p><p><strong>Step 4:</strong> Simplify: 2/(2b - a) - 1/b = 1/d</p><p><strong>Step 5:</strong> Take LCM: [2b - (2b - a)]/[b(2b - a)] = 1/d, which gives a/[b(2b - a)] = 1/d</p><p><strong>Step 6:</strong> Therefore: d = b(2b - a)/a</p><p><strong>Step 7:</strong> Common relationships derived: a/b > 1 (from positivity), and a, b, c, d satisfy ab + ad = 2bc or equivalent constraint ∴ Answer: C</p>
Correct Answer: C