<p><strong>139.</strong> If \(T_n\) denotes the \(n^{\text{th}}\) term of an arithmetic progression such that \(T_p=\dfrac{1}{q}\) and \(T_q=\dfrac{1}{p}\), then which of the given option is necessarily a root to the equation \((p+2q-3r)x^2+(q+2r-3p)x+(r+2p-3q)=0\), given that \(p+2q-3r\neq 0\)?</p>
Step-by-Step Solution
Key Concept: Use the AP property to find T_pq, then substitute into the quadratic. The cyclic symmetry of coefficients in the form (a, b, c) = (p+2q-3r, q+2r-3p, r+2p-3q) suggests testing x = T_pq as a root, which reveals that T_pq = 1/(pq) is necessarily satisfied.
<p><strong>Step 1: Set up AP relations</strong></p><p>For AP: T_n = a + (n-1)d</p><p>Given: T_p = 1/q and T_q = 1/p</p><p>So: a + (p-1)d = 1/q ... (i)</p><p>a + (q-1)d = 1/p ... (ii)</p><p><strong>Step 2: Find d and a</strong></p><p>Subtracting (i) from (ii):</p><p>(q-p)d = 1/p - 1/q = (q-p)/(pq)</p><p>∴ d = 1/(pq)</p><p>From (i): a = 1/q - (p-1)·1/(pq) = (p - p + 1)/(pq) = 1/(pq)</p><p><strong>Step 3: Find T_pq</strong></p><p>T_pq = a + (pq-1)d = 1/(pq) + (pq-1)·1/(pq) = pq/(pq) = 1</p><p><strong>Step 4: Verify x = 1 is a root</strong></p><p>Substitute x = 1 into the quadratic:</p><p>(p+2q-3r)(1)² + (q+2r-3p)(1) + (r+2p-3q)</p><p>= p + 2q - 3r + q + 2r - 3p + r + 2p - 3q</p><p>= (p - 3p + 2p) + (2q + q - 3q) + (-3r + 2r + r) = 0 ✓</p><p><strong>Alternatively, key insight:</strong> The cyclic structure (a+b+c = 0) in the coefficients guarantees x = 1 is always a root regardless of p, q, r values.</p><p>∴ <strong>Answer: x = 1</strong></p>
Correct Answer: A