<p><strong>308.</strong> If \(a_1, a_2, \ldots, a_n\) is a sequence of positive numbers which are in A.P. with common difference \(d\) and \(a_1 + a_4 + a_7 + \ldots + a_{16} = 147\) then \(a_1 + a_{16} = M\) and \(a_1 + a_6 + a_{11} + a_{16} = N\).</p><p>Maximum value of \(a_1 a_2 \ldots a_{16} = \left(\dfrac{S}{W}\right)^{16}\) (where \(S\) and \(W\) are coprime), then:</p>
Step-by-Step Solution
Key Concept: Recognize that a₁ + a₄ + a₇ + ... + a₁₆ forms an A.P. with 6 terms and common difference 3d. Use this to find M = a₁ + a₁₆, then apply AM-GM inequality on the 16 terms to maximize their product.
<p><strong>Step 1: Find M = a₁ + a₁₆</strong></p><p>The sequence a₁, a₄, a₇, ..., a₁₆ forms an A.P. with first term a₁, common difference 3d, and 6 terms.</p><p>Sum = a₁ + a₄ + a₇ + a₁₀ + a₁₃ + a₁₆ = 6a₁ + (0+3d+6d+9d+12d+15d) = 6a₁ + 45d = 147</p><p>Therefore: a₁ + 7.5d = 24.5</p><p>Also: a₁₆ = a₁ + 15d, so M = a₁ + a₁₆ = 2a₁ + 15d = <strong>49</strong></p><p><strong>Step 2: Find N = a₁ + a₆ + a₁₁ + a₁₆</strong></p><p>This sum of 4 equally spaced terms = 4a₁ + (0+5d+10d+15d) = 4a₁ + 30d = 2(2a₁ + 15d) = 2M = <strong>98</strong></p><p><strong>Step 3: Maximize a₁a₂...a₁₆ using AM-GM</strong></p><p>By AM-GM inequality: (a₁ + a₂ + ... + a₁₆)/16 ≥ ¹⁶√(a₁a₂...a₁₆)</p><p>Sum of 16 terms in A.P. = 8(a₁ + a₁₆) = 8M = 8(49) = 392</p><p>Average = 392/16 = 49/2</p><p>Maximum product = (49/2)¹⁶ occurs when all terms equal, i.e., a₁ = a₂ = ... = a₁₆ = 49/2</p><p>Therefore: S = 49, W = 2 (coprime), so answer options involve these values.</p><p>∴ Answer: <strong>ACD</strong></p>
Correct Answer: ACD