Let A={1,6,11,16, …} and B={9,16,23,30, …} be the sets consisting of the first 2025 terms of two arithmetic progressions. Then n(A ∪ B) is

Solution

Maths sir · 4 steps

Union is total minus overlap, and the overlap is itself an AP with common difference 35.

1. Last terms. In A the first term is 1 and the difference is 5, so the last term is 1 + 2024 × 5 = 10121. In B the first term is 9 and the difference is 7, so the last term is 9 + 2024 × 7 = 14177.
2. Common terms. The first number in both lists is 16. After that, common terms repeat every LCM(5, 7) = 35. So they are 16, 51, 86, …, and none can go past 10121, the last term of A.

   16 + 35k ≤ 10121 ⇒ k ≤ 10105/35 = 288.7

3. Count the overlap. k runs over 0, 1, …, 288, so n(A ∩ B) = 289.
4. So. n(A ∪ B) = 2025 + 2025 - 289 = 3761, which is option (d).

The idea to keep. Common terms of two APs form a new AP: start at the first shared term and step by the LCM of the two differences. Stop at the smaller of the two last terms.

Answer from the NTA final answer key. Solution by PrepWiser.