<p><strong>32.</strong> If the \(p\)th, \(q\)th, \(r\)th, and \(s\)th terms of an A.P. are in G.P., then \(p - q, q - r, r - s\) are in</p>
Step-by-Step Solution
Key Concept: If four terms of an A.P. are in G.P., use the A.P. formula aₙ = a + (n-1)d and the G.P. condition (common ratio is constant) to establish relationships between the position indices. The differences between indices will themselves form a sequence with a specific property.
<p><strong>Step 1:</strong> Let the A.P. have first term a and common difference d. Then:</p><p>aₚ = a + (p-1)d, aᵧ = a + (q-1)d, aᵣ = a + (r-1)d, aₛ = a + (s-1)d</p><p><strong>Step 2:</strong> Since aₚ, aᵧ, aᵣ, aₛ are in G.P., the common ratio is constant:</p><p>$$\frac{a_q}{a_p} = \frac{a_r}{a_q} = \frac{a_s}{a_r} = k$$ (say)</p><p><strong>Step 3:</strong> This means: aᵧ² = aₚ·aᵣ and aᵣ² = aᵧ·aₛ</p><p><strong>Step 4:</strong> Substituting A.P. terms and simplifying:</p><p>[a + (q-1)d]² = [a + (p-1)d]·[a + (r-1)d]</p><p>Expanding: a² + 2a(q-1)d + (q-1)²d² = a² + a[(p-1) + (r-1)]d + (p-1)(r-1)d²</p><p>This gives: 2(q-1) = (p-1) + (r-1), so 2q = p + r</p><p><strong>Step 5:</strong> Similarly, from aᵣ² = aᵧ·aₛ: 2r = q + s</p><p><strong>Step 6:</strong> From 2q = p + r, we get q - p = r - q, and from 2r = q + s, we get r - q = s - r</p><p>Let p - q = x, then q - r = x, and r - s = x</p><p>This means: $$\frac{1}{p-q}, \frac{1}{q-r}, \frac{1}{r-s}$$ are in A.P.</p><p><strong>Step 7:</strong> Therefore, (p-q), (q-r), (r-s) are in H.P. (Harmonic Progression)</p><p>∴ Answer: B (H.P.)</p>
Correct Answer: B