<p>If \(S_n = nP + \dfrac{n(n-1)}{2}Q\), where \(S_n\) denotes the sum of the first \(n\) terms of an A.P., then find the common difference.</p>
Step-by-Step Solution
Key Concept: For an A.P., the n-th term aₙ = Sₙ - Sₙ₋₁. Since Sₙ is quadratic in n, the common difference d is twice the coefficient of n² in the standard form Sₙ = An² + Bn.
<p><strong>Step 1:</strong> Rewrite Sₙ in standard quadratic form.</p><p>Sₙ = nP + n(n-1)Q/2 = nP + (Qn² - Qn)/2 = (Q/2)n² + (P - Q/2)n</p><p><strong>Step 2:</strong> For an A.P., Sₙ = (a/2)[2a + (n-1)d] = (d/2)n² + [a - d/2]n, where the coefficient of n² is d/2.</p><p><strong>Step 3:</strong> Comparing coefficients of n²: d/2 = Q/2, therefore d = Q.</p><p><strong>Step 4:</strong> Verify: From Sₙ = (Q/2)n² + (P - Q/2)n, the first term a = P and common difference d = Q. ✓</p><p>∴ Answer: Q</p>
Correct Answer: Q