<p>If \(a_1, a_2, \ldots, a_n\) are in HP, then the expression \(a_1a_2 + a_2a_3 + \cdots + a_{n-1}a_n\) is equal to</p>
Step-by-Step Solution
Key Concept: If a₁, a₂, ..., aₙ are in HP, then 1/a₁, 1/a₂, ..., 1/aₙ are in AP. Use this transformation to express the sum of products in terms of AP properties.
<p><strong>Step 1:</strong> Since a₁, a₂, ..., aₙ are in HP, let bᵢ = 1/aᵢ. Then b₁, b₂, ..., bₙ are in AP with first term b₁ and common difference d.</p><p><strong>Step 2:</strong> Express each term: bᵢ = b₁ + (i-1)d, so aᵢ = 1/[b₁ + (i-1)d]</p><p><strong>Step 3:</strong> The sum becomes: a₁a₂ + a₂a₃ + ... + aₙ₋₁aₙ = Σ(aᵢaᵢ₊₁) for i=1 to n-1</p><p><strong>Step 4:</strong> Substitute aᵢ = 1/bᵢ: Σ(1/(bᵢbᵢ₊₁)) = Σ(1/[bᵢ(bᵢ + d)])</p><p><strong>Step 5:</strong> Use partial fractions: 1/[bᵢ(bᵢ + d)] = (1/d)[1/bᵢ - 1/(bᵢ + d)] = (1/d)[1/bᵢ - 1/bᵢ₊₁]</p><p><strong>Step 6:</strong> This creates a telescoping series: (1/d)[1/b₁ - 1/bₙ] = (1/d)[a₁ - aₙ]</p><p>∴ Answer: D (or equivalent form depending on options given)</p>
Correct Answer: D