<p>Find the number of common terms in the following sequences: (ii) 1, 5, 9, … to 100 terms and 4, 7, 10, … to 60 terms.</p>
Step-by-Step Solution
Key Concept: Find the general terms of both APs, then determine when they're equal by solving for common terms. Use modular arithmetic or equate the general terms and find which values satisfy both sequence constraints.
<p><strong>Step 1:</strong> Write general terms of both sequences.</p><p>Sequence 1: a<sub>n</sub> = 1 + (n-1)·4 = 4n - 3, where 1 ≤ n ≤ 100</p><p>Sequence 2: b<sub>m</sub> = 4 + (m-1)·3 = 3m + 1, where 1 ≤ m ≤ 60</p><p><strong>Step 2:</strong> Equate the terms for common elements: 4n - 3 = 3m + 1</p><p>This gives: 4n = 3m + 4, so 4n ≡ 4 (mod 3)</p><p>Simplifying: n ≡ 1 (mod 3), meaning n = 3k + 1 for non-negative integers k</p><p><strong>Step 3:</strong> Substitute n = 3k + 1 into sequence 1: 4(3k + 1) - 3 = 12k + 1</p><p>This is the general form of common terms.</p><p><strong>Step 4:</strong> Apply constraints from both sequences:</p><p>From sequence 1: 1 ≤ 3k + 1 ≤ 100 → 0 ≤ k ≤ 33</p><p>From sequence 2: We need 12k + 1 = 3m + 1, so m = 4k. For sequence 2: 1 ≤ 4k ≤ 60 → 0.25 ≤ k ≤ 15</p><p><strong>Step 5:</strong> Combined constraint: 0 ≤ k ≤ 15</p><p>Therefore k ∈ {0, 1, 2, 3, ..., 15}</p><p>∴ <strong>Answer: 15</strong> (values k = 0, 1, 2, ..., 14 give 15 common terms)</p>
Correct Answer: 15