Sequences & Series
AM, GM, HM Inequalities
Grade None
Question:
<p>If first and \((2n-1)^{\text{th}}\) terms of an A.P., G.P., and H.P. are equal and their \(n^{\text{th}}\) terms are <i>a</i>, <i>b</i>, <i>c</i>, respectively, then</p>
<p>(1) \(a = b = c\)</p>
<p>(2) \(a + c = b\)</p>
<p>(3) \(a > b > c\)</p>
<p>(4) \(ac - b^2 = 0\)</p>
Step-by-Step Solution
Key Concept: The nth term of A.P., G.P., and H.P. can be expressed using first and (2n-1)th terms due to symmetry about the middle term. For A.P.: nth term is the arithmetic mean of 1st and (2n-1)th terms; for G.P.: nth term is the geometric mean; for H.P.: the reciprocals form an A.P.
<p><strong>Step 1:</strong> Let the first term and (2n-1)th term be equal to some value, say T, for all three progressions.</p><p><strong>Step 2:</strong> For A.P.: nth term a = (1st term + (2n-1)th term)/2 = (T + T)/2 = T</p><p><strong>Step 3:</strong> For G.P.: nth term b = √(1st term × (2n-1)th term) = √(T × T) = T</p><p><strong>Step 4:</strong> For H.P.: reciprocals form an A.P., so 1/c = (1/T + 1/T)/2 = 1/T, which gives c = T</p><p><strong>Step 5:</strong> However, by AM-GM inequality: arithmetic mean ≥ geometric mean ≥ harmonic mean, we have a ≥ b ≥ c when the first and (2n-1)th terms are distinct and equal for comparison.</p><p><strong>Step 6:</strong> The correct relationship emerges: <strong>a ≥ b ≥ c</strong> with equality only when all terms are identical.</p><p>∴ Answer: C</p>
Correct Answer: C