Sequences & Series
Arithmetic Progression
Grade 11

Question:

<p>Given an AP whose terms are all positive integers. The sum of its first nine terms is greater than 200 and less than 220. If the second term in it is 12, then its 4<sup>th</sup> term is</p>
<p>8</p>
<p>16</p>
<p>20</p>
<p>24</p>

Step-by-Step Solution

Key Concept: Use the AP sum formula S₉ = 9/2(2a + 8d) with constraint 200 < S₉ < 220 and a₂ = 12 to find d, then calculate a₄.
<p><strong>Step 1:</strong> Given a₂ = 12, so a + d = 12, which means a = 12 - d</p><p><strong>Step 2:</strong> Sum of first 9 terms: S₉ = 9/2(2a + 8d) = 9/2(2(12-d) + 8d) = 9/2(24 + 6d) = 9(12 + 3d) = 108 + 27d</p><p><strong>Step 3:</strong> Apply constraint: 200 < 108 + 27d < 220</p><p>⟹ 92 < 27d < 112</p><p>⟹ 3.407... < d < 4.148...</p><p>Since d must be a positive integer (for all terms to be positive integers), d = 4</p><p><strong>Step 4:</strong> Verify: S₉ = 108 + 27(4) = 108 + 108 = 216 ✓ (200 < 216 < 220)</p><p>Also a = 12 - 4 = 8 > 0 ✓</p><p><strong>Step 5:</strong> a₄ = a + 3d = 8 + 3(4) = 8 + 12 = 20</p><p>∴ Answer: D (20)</p>
Correct Answer: D

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