Sequences & Series
Arithmetic Progression
Grade 11
Question:
<p>The largest term common to the sequences 1, 11, 21, 31, … to 100 terms and 31, 36, 41, 46, … to 100 terms is</p>
<p>381</p>
<p>471</p>
<p>281</p>
<p>521</p>
Step-by-Step Solution
Key Concept: Find the general terms of both arithmetic progressions, then determine which terms are common by solving for when a_m = b_n, establishing that common terms form their own AP with common difference = LCM(10,5) = 10.
<p><strong>Step 1:</strong> Write general terms of both sequences.</p><p>First sequence: a_m = 1 + (m-1)·10 = 10m - 9, where m ≤ 100</p><p>Second sequence: b_n = 31 + (n-1)·5 = 5n + 26, where n ≤ 100</p><p><strong>Step 2:</strong> For common terms, set a_m = b_n.</p><p>10m - 9 = 5n + 26</p><p>10m = 5n + 35</p><p>2m = n + 7</p><p>n = 2m - 7</p><p><strong>Step 3:</strong> For valid common terms, both m and n must satisfy their respective bounds.</p><p>From n ≤ 100: 2m - 7 ≤ 100 → m ≤ 53.5 → m ≤ 53</p><p>From n ≥ 1: 2m - 7 ≥ 1 → m ≥ 4</p><p>So m ∈ {4, 5, 6, ..., 53}</p><p><strong>Step 4:</strong> Find the largest common term using m = 53.</p><p>a_53 = 10(53) - 9 = 530 - 9 = 521</p><p>Verify: n = 2(53) - 7 = 99, b_99 = 5(99) + 26 = 495 + 26 = 521 ✓</p><p>∴ Answer: <strong>521</strong></p>
Correct Answer: D