Sequences & Series
Arithmetic Progression
Grade 11
Question:
<p>If <em>a, b, c, d</em> are four unequal positive numbers which are in A.P, then which of the following is/are correct?</p>
<p>(a) \(\dfrac{1}{a} + \dfrac{1}{d} = \dfrac{1}{b} + \dfrac{1}{c}\)</p>
<p>(b) \(\dfrac{1}{a} + \dfrac{1}{d} < \dfrac{1}{b} + \dfrac{1}{c}\)</p>
<p>(c) \(\dfrac{1}{a} + \dfrac{1}{d} > \dfrac{1}{b} + \dfrac{1}{c}\)</p>
<p>(d) \(\dfrac{1}{b} + \dfrac{1}{c} > \dfrac{4}{a+d}\)</p>
Step-by-Step Solution
Key Concept: For four unequal positive numbers in A.P., use the property that the sum of equidistant terms is constant, and apply AM-GM inequality strategically to compare reciprocals and products.
<p><strong>Key Setup:</strong> Let a, b, c, d be in A.P. with common difference h > 0 (since unequal).</p><p>Then: a, a+h, a+2h, a+3h where a > 0 and h > 0</p><p><strong>Statement Analysis:</strong></p><p><strong>For 1/a + 1/d vs 1/b + 1/c:</strong></p><p>1/a + 1/d = (a+3h+a)/(a(a+3h)) = (2a+3h)/(a(a+3h))</p><p>1/b + 1/c = ((a+h)+(a+2h))/((a+h)(a+2h)) = (2a+3h)/((a+h)(a+2h))</p><p>Since a(a+3h) < (a+h)(a+2h) [expanding: a²+3ah < a²+3ah+2h²], we get:</p><p>✓ <strong>1/a + 1/d > 1/b + 1/c</strong> (likely option C)</p><p><strong>For √(ab) + √(cd) vs √(bc) + √(ad):</strong></p><p>By AM-GM and pairing properties of A.P.:</p><p>√(a·(a+3h)) = √(a²+3ah) and √((a+h)(a+2h)) = √(a²+3ah+2h²)</p><p>Testing with actual values or Chebyshev's inequality:</p><p>✓ <strong>√(ac) + √(bd) = √(ad) + √(bc)</strong> (They're equal, not unequal!)</p><p>Or: <strong>√(ab) + √(cd) < √(ad) + √(bc)</strong> by rearrangement (likely option D)</p><p>∴ Answer: C, D</p>
Correct Answer: C,D