<p>If \(H_1, H_2, \ldots, H_{20}\) are 20 harmonic means between 2 and 3, then \(\dfrac{H_1 + 2}{H_1 - 2} + \dfrac{H_{20} + 3}{H_{20} - 3} =\)</p>
Step-by-Step Solution
Key Concept: Harmonic means between two numbers form an HP (reciprocals form an AP). If H₁, H₂, ..., H₂₀ are HMs between 2 and 3, then 1/2, 1/H₁, 1/H₂, ..., 1/H₂₀, 1/3 form an AP with 22 terms. Use this to find relationships between H₁ and H₂₀.
<p><strong>Step 1:</strong> If H₁, H₂, ..., H₂₀ are 20 harmonic means between 2 and 3, then the sequence 1/2, 1/H₁, 1/H₂, ..., 1/H₂₀, 1/3 forms an AP with 22 terms.</p><p><strong>Step 2:</strong> Let d be the common difference. Then: 1/3 = 1/2 + 21d, so 21d = 1/3 - 1/2 = -1/6, giving d = -1/126.</p><p><strong>Step 3:</strong> First HM: 1/H₁ = 1/2 + d = 1/2 - 1/126 = (63-1)/126 = 62/126 = 31/63, so H₁ = 63/31.</p><p><strong>Step 4:</strong> Last HM: 1/H₂₀ = 1/2 + 20d = 1/2 - 20/126 = (63-20)/126 = 43/126, so H₂₀ = 126/43.</p><p><strong>Step 5:</strong> Calculate (H₁ + 2)/(H₁ - 2) = (63/31 + 2)/(63/31 - 2) = (63 + 62)/(63 - 62) = 125/1 = 125.</p><p><strong>Step 6:</strong> Calculate (H₂₀ + 3)/(H₂₀ - 3) = (126/43 + 3)/(126/43 - 3) = (126 + 129)/(126 - 129) = 255/(-3) = -85.</p><p><strong>Step 7:</strong> Sum = 125 + (-85) = 40.</p><p>∴ Answer: C</p>
Correct Answer: C