<p>Let \(a_1, a_2, a_3, \ldots, a_{49}\) be in A.P. such that \(\displaystyle\sum_{k=0}^{12} a_{4k+1} = 416\) and \(a_9 + a_{43} = 66\). If \(a_1^2 + a_2^2 + \ldots + a_{17}^2 = 140m\), then \(m\) is equal to</p>
Step-by-Step Solution
Key Concept: In an A.P., terms equidistant from ends are symmetric about the center. Use this to find the first term and common difference, then calculate the sum of squares of the first 17 terms using the formula for sum of squares in A.P.
<p><strong>Step 1:</strong> Identify the selected terms. ∑(k=0 to 12) a_(4k+1) selects: a₁, a₅, a₉, a₁₃, a₁₇, a₂₁, a₂₅, a₂₉, a₃₃, a₃₇, a₄₁, a₄₅, a₄₉ (13 terms forming an A.P. with common difference 4d)</p><p><strong>Step 2:</strong> Let a₁ = a and common difference = d. These 13 terms form an A.P. with first term a and common difference 4d. Sum = (13/2)(2a + 12·4d) = (13/2)(2a + 48d) = 416</p><p>Therefore: 13(a + 24d) = 416, so a + 24d = 32</p><p><strong>Step 3:</strong> Use the condition a₉ + a₄₃ = 66. We have a₉ = a + 8d and a₄₃ = a + 42d. Thus: 2a + 50d = 66, so a + 25d = 33</p><p><strong>Step 4:</strong> Solve the system: From Step 2 and Step 3, subtracting gives d = 1 and a = 8</p><p><strong>Step 5:</strong> Calculate a₁² + a₂² + ... + a₁₇². The terms are 8, 9, 10, ..., 24. Use the formula: ∑(k=1 to 17) (7+k)² = ∑(k=1 to 17) (49 + 14k + k²) = 17(49) + 14·(17·18/2) + (17·18·35/6) = 833 + 2142 + 1785 = 4760</p><p><strong>Step 6:</strong> Given a₁² + a₂² + ... + a₁₇² = 140m, we have 4760 = 140m, so m = 34</p><p>∴ Answer: D (m = 34)</p>
Correct Answer: D