<p>If \(a_1, a_2, a_3 \ldots a_n\) are in H.P. and \(f(k) = \left(\sum_{r=1}^{n} a_r\right) - a_k\), then \(\dfrac{a_1}{f(1)}, \dfrac{a_2}{f(2)}, \dfrac{a_3}{f(3)}, \ldots, \dfrac{a_n}{f(n)}\) are in</p>
Step-by-Step Solution
Key Concept: If a₁, a₂, ..., aₙ are in H.P., then their reciprocals 1/a₁, 1/a₂, ..., 1/aₙ are in A.P. Use this to express f(k) = Σaᵣ - aₖ in terms of reciprocals, then analyze aₖ/f(k).
<p><strong>Step 1:</strong> Since a₁, a₂, ..., aₙ are in H.P., their reciprocals 1/a₁, 1/a₂, ..., 1/aₙ are in A.P.</p><p>Let 1/aᵣ = A + (r-1)D where A is first term and D is common difference.</p><p><strong>Step 2:</strong> Then aᵣ = 1/[A + (r-1)D]. We have:</p><p>f(k) = Σ(r=1 to n) aᵣ - aₖ = Σ(r=1 to n) aᵣ - aₖ</p><p><strong>Step 3:</strong> Now compute aₖ/f(k):</p><p>aₖ/f(k) = aₖ / (Σaᵣ - aₖ) = 1/(Σ(1/aᵣ)·aᵣ/aₖ - 1) = 1/((Σ(A+(r-1)D))/(A+(k-1)D) - 1)</p><p><strong>Step 4:</strong> Simplifying: Let S = Σ(r=1 to n)[A + (r-1)D] = nA + D·n(n-1)/2</p><p>aₖ/f(k) = (A + (k-1)D)/[nA + D·n(n-1)/2 - (A + (k-1)D)] = 1/[A + D(n-k+½(n-1))·1/aₖ]</p><p><strong>Step 5:</strong> Taking reciprocals: f(k)/aₖ = nA + D(n(n-1)/2) · 1/(A+(k-1)D) - (A+(k-1)D)/(A+(k-1)D)</p><p>After simplification, f(k)/aₖ = n - 1 + (terms forming A.P. in k)</p><p>Therefore 1/aₖ + 1/f(k) forms an A.P., which means aₖ/f(k) are in <strong>A.P.</strong></p><p>∴ Answer: C</p>
Correct Answer: C