<p>If P = (1/x_p, p); Q = (1/x_q, q); R = (1/x_r, r) where x_k ≠ 0, denotes the k-th terms of a H.P. for k ∈ ℕ, then:</p>
<p>(a) ar(ΔPQR) = (p² + q² + r²)/2 - (p-q)² - (q-r)² - (r-p)²</p>
<p>(b) ΔPQR is a right angled triangle</p>
<p>(c) the points P, Q, R are collinear</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: If the k-th terms of a H.P. form an A.P. in their reciprocals, then the points with coordinates (1/xₖ, k) must lie on a straight line since the reciprocals form an A.P., making the first coordinate form a linear relationship with the second coordinate.
<p><strong>Step 1:</strong> Recall that k-th term of a H.P. is xₖ = 1/aₖ where {aₖ} is an A.P.</p><p><strong>Step 2:</strong> If xₚ, xᵩ, xᵣ are terms of a H.P., then 1/xₚ, 1/xᵩ, 1/xᵣ form an A.P.</p><p><strong>Step 3:</strong> Let 1/xₚ = a, 1/xᵩ = b, 1/xᵣ = c. Since these form an A.P., we have: b - a = c - b, or equivalently, 2b = a + c.</p><p><strong>Step 4:</strong> The three points are P = (a, p), Q = (b, q), R = (c, r) where a, b, c form an A.P. with corresponding y-coordinates p, q, r.</p><p><strong>Step 5:</strong> For collinearity, check the slope condition. Slope PQ = (q-p)/(b-a) and Slope QR = (r-q)/(c-b). Since the H.P. property makes 1/xₖ form an A.P., the natural spacing means these points lie on a line of form: y = mx + n, where y-coordinate is linearly related to the sequence index.</p><p><strong>Step 6:</strong> The points P, Q, R with coordinates (1/xₖ, k) where {1/xₖ} forms an A.P. and the y-coordinates are p, q, r (which are consecutive indices or related to sequence position) must be collinear.</p><p><strong>Step 7:</strong> Using the determinant test for collinearity: |a(q-r) + b(r-p) + c(p-q)| = 0. Since a, b, c form an A.P. (i.e., b-a = c-b) and if p, q, r correspond to equally spaced positions, the determinant equals zero.</p><p><strong>∴ Answer: C</strong></p>
Correct Answer: C