<p>If \(a, b, c, d \in R^+\) and \(a, b, c, d\) are in H.P., then</p>
<p>(1) \(a + d > b + c\)</p>
<p>(2) \(a + b > c + d\)</p>
<p>(3) \(a + c > b + d\)</p>
<p>(4) none of these</p>
Step-by-Step Solution
Key Concept: If a, b, c, d are in H.P., then their reciprocals 1/a, 1/b, 1/c, 1/d form an A.P. Use this transformation along with properties of A.P. and inequalities to establish relationships between the terms.
<p><strong>Step 1:</strong> Since a, b, c, d are in H.P., their reciprocals are in A.P.</p><p>So: 1/a, 1/b, 1/c, 1/d are in A.P.</p><p><strong>Step 2:</strong> For an A.P., common difference is constant:</p><p>1/b - 1/a = 1/c - 1/b = 1/d - 1/c = k (say)</p><p><strong>Step 3:</strong> This gives us: 1/b = 1/a + k, 1/c = 1/a + 2k, 1/d = 1/a + 3k</p><p><strong>Step 4:</strong> Key relationship: Since a, b, c, d ∈ ℝ⁺ and form H.P., we have:</p><p>• a < b < c < d (if k > 0, i.e., decreasing H.P.)</p><p>• a + d = b + c is NOT true for H.P.</p><p>• Instead: 1/a + 1/d = 1/b + 1/c (equidistant terms in A.P. have equal reciprocals)</p><p><strong>Step 5:</strong> From reciprocal property: a + d > b + c (for H.P. in natural order)</p><p>Or examine: 1/(b+c) < 1/(a+d) which means b+c > a+d</p><p>∴ The most common statement is: <strong>a + d > b + c</strong> or equivalently relationships derived from A.P. of reciprocals</p>
Correct Answer: A