Let S₁ and S₂ be two sets containing the first 200 terms of the arithmetic progressions 3, 7, 11, and 4, 9, 14, respectively. The number of elements in S₁ S₂ that are not divisible by 3 is

Step-by-step solution

The set S₁ consists of 200 terms with a=3, d=4 . The last term is 3 + 199 4 = 799 . The set S₂ consists of 200 terms with a=4, d=5 . The last term is 4 + 199 5 = 999 . First, we find the number of common terms in S₁ and S₂ . The common terms form an AP with common difference d = lcm (4, 5) = 20 . The first common term is 19 . Let the number of terms in S₁ S₂ be n . Then 19 + (n-1)20 799 20(n-1) 780 n-1 39 n 40 . So, n(S₁ S₂) = 40 . Total number of unique elements in S₁ S₂ is n(S₁) + n(S₂) - n(S₁ S₂) = 200 + 200 - 4