Let set A consist of the first 120 terms of the arithmetic progression 2, 5, 8, and set B consist of the first 100 terms of the arithmetic progression 3, 7, 11, . The number of elements in the set A B that are NOT divisible by 3 is

Step-by-step solution

Let us analyze the terms in set A and set B . For set A : The general term is a_n = 2 + (n-1)3 = 3n - 1 , where 1 n 120 . The largest term in A is 3(120) - 1 = 359 . Since every term is of the form 3n - 1 , no term in A is divisible by 3 . Thus, all 120 terms of A are not divisible by 3 . For set B : The general term is b_m = 3 + (m-1)4 = 4m - 1 , where 1 m 100 . The largest term in B is 4(100) - 1 = 399 . We need to find how many terms in B are divisible by 3 . Setting 4m - 1 0 3 , we get m - 1 0 3 , which means m