Sequences & Series
Arithmetic Progression
Grade 11
Question:
<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 options 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>
<p>(a) \(T_{pq}\)</p>
<p>(b) \(T_p\)</p>
<p>(c) \(T_q\)</p>
<p>(d) \(T_{p+q}\)</p>
Step-by-Step Solution
Key Concept: Use the AP properties to find the common difference and first term, then recognize that T_{pq} = 1 always, which must satisfy the given quadratic equation when x = T_{pq}.
<p><strong>Step 1:</strong> For AP with first term a and common difference d:</p><p>T_p = a + (p-1)d = 1/q ... (i)</p><p>T_q = a + (q-1)d = 1/p ... (ii)</p><p><strong>Step 2:</strong> Subtract (i) from (ii):</p><p>(q-p)d = 1/p - 1/q = (q-p)/(pq)</p><p>Therefore: d = 1/(pq)</p><p><strong>Step 3:</strong> Substitute back into (i):</p><p>a + (p-1)·1/(pq) = 1/q</p><p>a = 1/q - (p-1)/(pq) = (p - p + 1)/(pq) = 1/(pq)</p><p><strong>Step 4:</strong> Find T_{pq}:</p><p>T_{pq} = a + (pq-1)d = 1/(pq) + (pq-1)·1/(pq)</p><p>T_{pq} = [1 + pq - 1]/(pq) = 1</p><p><strong>Step 5:</strong> Verify that x = 1 satisfies the equation:</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>∴ Answer: A (x = 1 is a root, which equals T_{pq})</strong></p>
Correct Answer: A