<p>If \(x_1, x_2, \ldots, x_{20}\) are in H.P. and \(x_1, 2, x_{20}\) are in G.P., then \(\sum_{r=1}^{19} x_r x_{r+1} =\)</p>
Step-by-Step Solution
Key Concept: Since x₁, x₂, ..., x₂₀ are in H.P., their reciprocals form an A.P. Use the G.P. condition x₁, 2, x₂₀ to establish that the reciprocals form a symmetric A.P., then compute the sum using the property that consecutive terms in H.P. satisfy a specific relationship.
<p><strong>Step 1:</strong> Let 1/x₁, 1/x₂, ..., 1/x₂₀ form an A.P. with first term a and common difference d.</p><p>Then: 1/xᵣ = a + (r-1)d, so xᵣ = 1/(a + (r-1)d)</p><p><strong>Step 2:</strong> From x₁, 2, x₂₀ in G.P.: x₁·x₂₀ = 4</p><p>Also: 1/x₁ = a and 1/x₂₀ = a + 19d</p><p>Since x₁·x₂₀ = 4: (1/a)·(1/(a+19d)) = 4, which gives a(a+19d) = 1/4</p><p><strong>Step 3:</strong> From the G.P. condition: 2² = x₁·x₂₀, and since 2 must be the geometric mean: x₁·x₂₀ = 4</p><p>The reciprocals satisfy: 1/x₁ + 1/x₂₀ = 2a + 19d. For symmetry in H.P.: a + (a+19d) = 2·(1/2) = 1</p><p>Therefore: x₁ + x₂₀ = 4</p><p><strong>Step 4:</strong> For H.P. terms: xᵣ·xᵣ₊₁ = 1/[(a+(r-1)d)(a+rd)]</p><p>∑ᵣ₌₁¹⁹ xᵣxᵣ₊₁ = ∑ᵣ₌₁¹⁹ 1/[(a+(r-1)d)(a+rd)] = (1/d)·∑ᵣ₌₁¹⁹ [1/(a+(r-1)d) - 1/(a+rd)]</p><p><strong>Step 5:</strong> This telescopes: = (1/d)·[1/a - 1/(a+19d)] = (1/d)·(19d)/(a(a+19d)) = 19/(a(a+19d)) = 19·4 = 76</p><p>∴ Answer: <strong>A) 76</strong></p>
Correct Answer: A