<p>If \(a_1, a_2, \ldots, a_n\) are in H.P., then \(\dfrac{a_1}{a_2 + a_3 + \cdots + a_n},\ \dfrac{a_2}{a_1 + a_3 + \cdots + a_n},\ \ldots,\ \dfrac{a_n}{a_1 + a_2 + \cdots + a_{n-1}}\) are in</p>
Step-by-Step Solution
Key Concept: If a₁, a₂, ..., aₙ are in H.P., then their reciprocals form an A.P. Use this to show that the given expressions, when reciprocals are taken, form an A.P.
<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/aᵢ = b₁ + (i-1)d for some common difference d.</p><p><strong>Step 2:</strong> Consider the reciprocal of the first term of the given sequence:</p><p>$$\frac{a_2 + a_3 + \cdots + a_n}{a_1} = \frac{1}{a_1}\left(\frac{a_2 + a_3 + \cdots + a_n}{a_1}\right) = \frac{1}{a_1}\sum_{i=2}^{n}a_i$$</p><p>$$= \frac{1}{a_1}\left(S - a_1\right)$$ where S = a₁ + a₂ + ... + aₙ</p><p><strong>Step 3:</strong> The reciprocal of the kth term is:</p><p>$$\frac{a_1 + a_2 + \cdots + a_{k-1} + a_{k+1} + \cdots + a_n}{a_k} = \frac{S - a_k}{a_k} = \frac{S}{a_k} - 1$$</p><p><strong>Step 4:</strong> Since S is constant and 1/aₖ forms an A.P., the reciprocals of our given terms form an A.P., which means the original terms are in H.P.</p><p>∴ Answer: <strong>A (Harmonic Progression)</strong></p>
Correct Answer: A