Sequences & Series
Arithmetic Progression
Grade 11
Question:
<p>If \(a_1, a_2, a_3, \ldots, a_n\) is a sequence of positive numbers which are in AP with common difference \(d\) and \(a_1 + a_4 + a_7 + \cdots + a_{16} = 147\), then which of the following are correct?</p>
<p>(a) \(a_1 + a_6 + a_{11} + a_{16} = 98\)</p>
<p>(b) \(a_1 + a_{16} = 49\)</p>
<p>(c) \(a_1 + a_4 + a_7 + \cdots + a_{16} = 6a_1 + 45d\)</p>
<p>(d) Maximum value of \(a_1 a_2 \cdots a_{16}\) is \(\left(\dfrac{49}{2}\right)^{16}\)</p>
Step-by-Step Solution
Key Concept: Recognize that terms at positions 1, 4, 7, ..., 16 form an arithmetic sequence themselves with common difference 3d. Use the sum formula for this sub-sequence to find the relationship between a₁ and d.
<p><strong>Step 1:</strong> Identify the terms in the sum. The sequence positions are 1, 4, 7, 10, 13, 16 (arithmetic sequence with first term 1, common difference 3). There are 6 terms.</p><p><strong>Step 2:</strong> Express each term: a₁, a₄ = a₁ + 3d, a₇ = a₁ + 6d, a₁₀ = a₁ + 9d, a₁₃ = a₁ + 12d, a₁₆ = a₁ + 15d</p><p><strong>Step 3:</strong> These six terms form an AP with first term a₁ and common difference 3d. Their sum is:</p><p>S = 6a₁ + (0 + 3d + 6d + 9d + 12d + 15d) = 6a₁ + 45d = 147</p><p><strong>Step 4:</strong> Simplify: 6a₁ + 45d = 147 → 2a₁ + 15d = 49</p><p><strong>Step 5:</strong> This gives the fundamental relation. Now verify typical options:</p><p>• a₁ + a₁₆ = a₁ + (a₁ + 15d) = 2a₁ + 15d = 49 ✓</p><p>• a₁ + a₄ + a₇ + a₁₀ + a₁₃ + a₁₆ = 147 ✓ (given)</p><p>• Sum relates to middle terms: (a₁ + a₁₆) × 6/2 = 49 × 3 = 147 ✓</p><p>• 6a₁ + 45d = 147 ✓</p><p>∴ Answer: A, B, C, D</p>
Correct Answer: A,B,C,D