Sequences & Series
Arithmetic Progression
Grade 11
Question:
<p>Let \(a_1, a_2, a_3, \ldots, a_n\) be in AP. If \(a_3 + a_7 + a_{11} + a_{15} = 72\), then the sum of its first 17 terms is equal to</p>
<p>306</p>
<p>204</p>
<p>153</p>
<p>612</p>
Step-by-Step Solution
Key Concept: In an AP, terms equidistant from the ends sum to the same value. Recognize that a₃ + a₁₅ = a₇ + a₁₁ = 2·(middle term of AP), and use S₁₇ = (17/2)·(first term + last term) = (17/2)·2·(middle term).
<p><strong>Step 1:</strong> Write terms in AP: aₙ = a₁ + (n-1)d</p><p><strong>Step 2:</strong> Express given condition: a₃ + a₇ + a₁₁ + a₁₅ = 72</p><p>(a₁ + 2d) + (a₁ + 6d) + (a₁ + 10d) + (a₁ + 14d) = 72</p><p><strong>Step 3:</strong> Simplify: 4a₁ + 32d = 72 → a₁ + 8d = 18</p><p><strong>Step 4:</strong> Recognize that a₁ + 8d is the 9th term (middle term of first 17 terms): a₉ = 18</p><p><strong>Step 5:</strong> Apply sum formula: S₁₇ = (17/2)(a₁ + a₁₇) = (17/2)(2a₉) = 17·a₉</p><p><strong>Step 6:</strong> Calculate: S₁₇ = 17 × 18 = 306</p><p>∴ Answer: A</p>
Correct Answer: A