<p>Let \(A_1, A_2\); \(G_1, G_2\) and \(H_1, H_2\) be two AM's, GM's and HM's respectively between two positive real numbers \(a\) and \(b\), then:</p>
<p>(a) \(A_1 H_2 = ab\)</p>
<p>(b) \(A_1 H_2 = a^2 b^2\)</p>
<p>(c) \(G_1 G_2 = ab\)</p>
<p>(d) \(A_2 H_1 = ab\)</p>
Step-by-Step Solution
Key Concept: When two AMs, GMs, and HMs are inserted between a and b, they form arithmetic, geometric, and harmonic progressions respectively. The relationship between these means follows from the property that A·H = G² for corresponding terms in the sequence.
<p><strong>Step 1:</strong> For two AMs A₁, A₂ between a and b, the sequence is a, A₁, A₂, b (arithmetic progression with 4 terms).</p><p>Common difference d = (b-a)/3</p><p>So A₁ = a + (b-a)/3 = (2a+b)/3 and A₂ = a + 2(b-a)/3 = (a+2b)/3</p><p><strong>Step 2:</strong> For two GMs G₁, G₂ between a and b, the sequence is a, G₁, G₂, b (geometric progression with 4 terms).</p><p>Common ratio r = (b/a)^(1/3)</p><p>So G₁ = a·r = a·(b/a)^(1/3) and G₂ = a·r² = a·(b/a)^(2/3)</p><p><strong>Step 3:</strong> For two HMs H₁, H₂ between a and b, the reciprocals form an AP.</p><p>1/a, 1/H₁, 1/H₂, 1/b is an AP with common difference d' = (1/b - 1/a)/3</p><p>So H₁ = 3ab/(2b+a) and H₂ = 3ab/(a+2b)</p><p><strong>Step 4:</strong> Verify the key relationship: A₁·A₂ = (2a+b)(a+2b)/9 and G₁·G₂ = a·b, H₁·H₂ = 9a²b²/(2b+a)(a+2b)</p><p>The fundamental relation is: <strong>A₁·A₂ = G₁² and H₁·H₂ relates through A·H = G²</strong></p><p>∴ Answer: C</p>
Correct Answer: C