<p>Let \(a_1, a_2, a_3, \ldots, a_{49}\) be in AP such that \(\displaystyle\sum_{k=0}^{12} a_{4k+1} = 416\) and \(a_9 + a_{43} = 66\). If \(a_1^2 + a_2^2 + \cdots + a_{17}^2 = 140m\), then <i>m</i> is equal to</p>
Step-by-Step Solution
Key Concept: Use the AP properties: (1) express the sum of specific terms using the general form a_n = a_1 + (n-1)d, (2) use the symmetry property that a_9 + a_43 = 66 to find the common difference, and (3) calculate the sum of squares of the first 17 terms.
<p><strong>Step 1: Set up the AP</strong></p><p>Let the AP be: a_n = a_1 + (n-1)d where a_1 is the first term and d is the common difference.</p><p><strong>Step 2: Use the condition a_9 + a_43 = 66</strong></p><p>a_9 = a_1 + 8d and a_43 = a_1 + 42d</p><p>a_9 + a_43 = 2a_1 + 50d = 66</p><p>∴ a_1 + 25d = 33 ... (1)</p><p><strong>Step 3: Analyze the sum ∑_{k=0}^{12} a_{4k+1}</strong></p><p>The terms are: a_1, a_5, a_9, a_13, ..., a_49 (13 terms in total, k goes from 0 to 12)</p><p>a_{4k+1} = a_1 + 4kd</p><p>∑_{k=0}^{12} a_{4k+1} = 13a_1 + 4d(0+1+2+...+12) = 13a_1 + 4d·(12·13/2) = 13a_1 + 312d = 416</p><p>∴ a_1 + 24d = 32 ... (2)</p><p><strong>Step 4: Solve for a_1 and d</strong></p><p>From (1): a_1 + 25d = 33</p><p>From (2): a_1 + 24d = 32</p><p>Subtracting: d = 1</p><p>Substituting back: a_1 = 32 - 24(1) = 8</p><p><strong>Step 5: Calculate ∑_{i=1}^{17} a_i²</strong></p><p>a_i = 8 + (i-1)·1 = 7 + i</p><p>∑_{i=1}^{17} a_i² = ∑_{i=1}^{17} (7+i)²</p><p>Let j = 7 + i, so when i goes from 1 to 17, j goes from 8 to 24</p><p>∑_{i=1}^{17} a_i² = ∑_{j=8}^{24} j² = ∑_{j=1}^{24} j² - ∑_{j=1}^{7} j²</p><p>Using ∑_{j=1}^{n} j² = n(n+1)(2n+1)/6:</p><p>∑_{j=1}^{24} j² = 24·25·49/6 = 4900</p><p>∑_{j=1}^{7} j² = 7·8·15/6 = 140</p><p>∑_{i=1}^{17} a_i² = 4900 - 140 = 4760</p><p><strong>Step 6: Find m</strong></p><p>Given: a_1² + a_2² + ... + a_17² = 140m</p><p>4760 = 140m</p><p>m = 4760/140 = 34</p><p><strong>∴ Answer: B</strong></p>
Correct Answer: B