<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 positive reals a and b, they form sequences with fixed relationships: the AMs satisfy A₁ + A₂ = a + b, the GMs satisfy G₁G₂ = ab, and the HMs are related by the reciprocal AM property. The key is recognizing that products and sums of these means follow specific patterns.
<p><strong>Step 1:</strong> For two AMs A₁, A₂ between a and b, the sequence a, A₁, A₂, b forms an AP.</p><p>Common difference: d = (b-a)/3</p><p>Therefore: 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 a, G₁, G₂, b forms a GP.</p><p>Common ratio: r = (b/a)^(1/3)</p><p>Therefore: G₁ = a·r = a^(2/3)·b^(1/3) and G₂ = a·r² = a^(1/3)·b^(2/3)</p><p>Product: G₁·G₂ = a^(2/3)·b^(1/3)·a^(1/3)·b^(2/3) = ab</p><p><strong>Step 3:</strong> For two HMs H₁, H₂ between a and b, the reciprocals form an AP: 1/a, 1/H₁, 1/H₂, 1/b.</p><p>Sum: H₁ + H₂ = 2ab/(a+b)</p><p><strong>Step 4:</strong> Verify key relationship: A₁ + A₂ = a + b, G₁G₂ = ab, and H₁H₂ = a²b²/(a+b)²</p><p><strong>Step 5:</strong> The fundamental relationship is: G₁² = A₁H₁ and G₂² = A₂H₂ (GM is geometric mean of corresponding AM and HM)</p><p>∴ Answer: C</p>
Correct Answer: C