<p>If \(a_1, a_2, a_3, \ldots\) are in H.P. and \(f(k) = \left(\displaystyle\sum_{r=1}^{n} a_r\right) - a_k\), 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, ... are in H.P., then 1/a₁, 1/a₂, ... are in A.P. Recognizing that f(k) = Σaᵣ - aₖ represents the sum of all terms except aₖ allows us to express aₖ/f(k) in terms of reciprocals, which reveals the pattern.
<p><strong>Step 1:</strong> Since a₁, a₂, a₃, ... are in H.P., we know 1/a₁, 1/a₂, 1/a₃, ... are in A.P.</p><p>Let 1/aᵣ = α + (r-1)d for some constants α and d.</p><p><strong>Step 2:</strong> Let S = Σ(r=1 to n) aᵣ. Then f(k) = S - aₖ.</p><p><strong>Step 3:</strong> Now consider aₖ/f(k). We need to find the reciprocal: f(k)/aₖ = (S - aₖ)/aₖ = S/aₖ - 1.</p><p><strong>Step 4:</strong> Express this as: f(k)/aₖ = S·(1/aₖ) - 1 = S(α + (k-1)d) - 1.</p><p><strong>Step 5:</strong> Since 1/aₖ forms an A.P., we have:</p><p>f(k)/aₖ = S·α + S·d·(k-1) - 1 = (Sα - 1) + Sd(k-1)</p><p>This is linear in k, meaning f(k)/aₖ is in A.P.</p><p><strong>Step 6:</strong> Therefore, aₖ/f(k) (the reciprocals) are in H.P.</p><p>∴ Answer: A (Harmonic Progression)
Correct Answer: A