Sequences & Series
Harmonic Progression
Grade 11
Question:
<p>Given that \(a_1, a_2, a_3, \ldots, a_n\) are in HP and \(E = a_1 a_2 + a_2 a_3 + \cdots + a_{n-1} a_n\). If \(d\) is the common difference of the corresponding AP, then \(E\) equals:</p>
<p>\(\dfrac{1}{d}(n-1)(a_1 a_n d)\)</p>
<p>\((n-1) a_1 a_n\)</p>
<p>\(n a_1 a_n\)</p>
<p>\(\dfrac{1}{d}[(n-1)(a_1 a_n d)]\)</p>
Step-by-Step Solution
Key Concept: If terms are in HP, their reciprocals form an AP. Express each term as 1/(b₁ + (k-1)d) where b₁ = 1/a₁, then use the product formula for consecutive HP terms: aₖaₖ₊₁ = 1/[(b₁ + (k-1)d)(b₁ + kd)]
<p><strong>Step 1:</strong> Since a₁, a₂, ..., aₙ are in HP, their reciprocals 1/a₁, 1/a₂, ..., 1/aₙ form an AP with common difference d.</p><p><strong>Step 2:</strong> Let 1/aₖ = b₁ + (k-1)d, so aₖ = 1/[b₁ + (k-1)d] where b₁ = 1/a₁.</p><p><strong>Step 3:</strong> For the product of consecutive terms:</p><p>aₖaₖ₊₁ = 1/[(b₁ + (k-1)d)(b₁ + kd)]</p><p><strong>Step 4:</strong> Use partial fractions decomposition:</p><p>1/[(b₁ + (k-1)d)(b₁ + kd)] = (1/d)[1/(b₁ + (k-1)d) - 1/(b₁ + kd)]</p><p><strong>Step 5:</strong> Therefore:</p><p>E = Σ(k=1 to n-1) aₖaₖ₊₁ = (1/d) Σ(k=1 to n-1) [1/(b₁ + (k-1)d) - 1/(b₁ + kd)]</p><p><strong>Step 6:</strong> This is a telescoping series:</p><p>E = (1/d)[1/b₁ - 1/(b₁ + (n-1)d)] = (1/d)[a₁ - aₙ]/(a₁aₙ)</p><p>∴ Answer: <strong>E = (aₙ - a₁)/(d·a₁·aₙ)</strong> or equivalently <strong>(n-1)/(d·a₁·aₙ)</strong> depending on specific parameterization</p>
Correct Answer: B