Sequences & Series
Arithmetic Progression
Grade 11
Question:
<p>If \(a, b, c, d\) are four unequal positive numbers which are in A.P., then</p>
<p>\(\dfrac{1}{a} + \dfrac{1}{d} > \dfrac{1}{b} + \dfrac{1}{c}\)</p>
<p>\(\dfrac{1}{a} + \dfrac{1}{d} < \dfrac{1}{b} + \dfrac{1}{c}\)</p>
<p>\(\dfrac{1}{b} + \dfrac{1}{c} > \dfrac{4}{a+d}\)</p>
<p>\(\dfrac{1}{a} + \dfrac{1}{d} = \dfrac{1}{b} + \dfrac{1}{c}\)</p>
Step-by-Step Solution
Key Concept: For numbers in A.P., use the property that b-a = c-b = d-c = k (common difference). Recognize that reciprocals of A.P. terms form a specific pattern that can be compared using AM-HM inequality or direct algebraic manipulation.
<p><strong>Step 1:</strong> Let a, b, c, d be in A.P. with common difference k > 0 (since they're unequal).</p><p>Then: b = a+k, c = a+2k, d = a+3k</p><p><strong>Step 2:</strong> We need to compare 1/a + 1/d with 1/b + 1/c</p><p>1/a + 1/d = 1/a + 1/(a+3k) = (a+3k+a)/[a(a+3k)] = (2a+3k)/[a(a+3k)]</p><p><strong>Step 3:</strong> 1/b + 1/c = 1/(a+k) + 1/(a+2k) = (a+2k+a+k)/[(a+k)(a+2k)] = (2a+3k)/[(a+k)(a+2k)]</p><p><strong>Step 4:</strong> Compare denominators: a(a+3k) vs (a+k)(a+2k)</p><p>a(a+3k) = a² + 3ak</p><p>(a+k)(a+2k) = a² + 3ak + 2k²</p><p><strong>Step 5:</strong> Since a, k > 0, we have a(a+3k) < (a+k)(a+2k)</p><p>Therefore: 1/a + 1/d > 1/b + 1/c</p><p>∴ Answer: AC</p>
Correct Answer: AC