Sequences & Series
AM, GM, HM
Grade None
Question:
<p>If \(A_1, A_2, G_1, G_2\) and \(H_1, H_2\) are two arithmetic, geometric and harmonic means, respectively, between two quantities \(a\) and \(b\), then \(ab\) is equal to</p>
<p>\(A_1 H_2\)</p>
<p>\(A_2 H_1\)</p>
<p>\(G_1 G_2\)</p>
<p>none of these</p>
Step-by-Step Solution
Key Concept: When two AMs, GMs, and HMs are inserted between a and b, they create specific sequences where the product of the means relates to a·b through the property that geometric means equal √(ab) and harmonic means satisfy 1/H = arithmetic mean of 1/a and 1/b.
<p><strong>Step 1:</strong> For two AMs A₁, A₂ between a and b: The sequence is a, A₁, A₂, b with common difference d = (b-a)/3</p><p><strong>Step 2:</strong> For two GMs G₁, G₂ between a and b: G₁·G₂ = ab (property of geometric means)</p><p><strong>Step 3:</strong> For two HMs H₁, H₂ between a and b: Their reciprocals 1/H₁, 1/H₂ form an AP between 1/a and 1/b</p><p><strong>Step 4:</strong> The sequence of reciprocals is 1/a, 1/H₁, 1/H₂, 1/b with common difference d' = (1/b - 1/a)/3</p><p><strong>Step 5:</strong> This gives 1/H₁ = 1/a + (1/b - 1/a)/3 and 1/H₂ = 1/a + 2(1/b - 1/a)/3</p><p><strong>Step 6:</strong> Multiplying: (1/H₁)·(1/H₂) = [1/a + (1/b - 1/a)/3]·[1/a + 2(1/b - 1/a)/3] = 1/(ab)</p><p><strong>Step 7:</strong> Therefore H₁·H₂ = ab</p><p>∴ Answer: C (ab equals H₁·H₂)</p>
Correct Answer: C