Let X be the set consisting of the first 2018 term of the arithmetic progression 1, 6, 11, .………… , and Y ‘be the set consisting of the first 2018 terms of the arithmetic progression 9, 16, 23, ……….. Then, the number of elements in the set X∪Y is 

Let X be the set consisting of the first 2018 term of the arithmetic progression 1, 6, 11, .………… , and Y 'be the set consisting of the first 2018 terms of the arithmetic progression 9, 16, 23, ……….. Then, the number of elements in the set XY is 

    Register to Get Free Mock Test and Study Material

    +91

    Verify OTP Code (required)

    I agree to the terms and conditions and privacy policy.

    Solution:

    Here X={1,6,11,,10086}

    and Y={9,16,23,,14128}

    The intersection of X and Y is an A.P. with 16 as first term and 35 as common difference.

    The series becomes 16, 51, 86, ..

    Now, kth  term =16+(k1)3510086

    i.e. k1010535k288 (as k is to be an integer)

    Hence,n(XY)=n(X)+n(Yn(XY)=2018+2018288=3748

    Chat on WhatsApp Call Infinity Learn

      Register to Get Free Mock Test and Study Material

      +91

      Verify OTP Code (required)

      I agree to the terms and conditions and privacy policy.