Discussion Forum

Let X  be the set consisting of the first 2018 terms 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 υ Y is.....

Answer:

Explanation:

Here X= {1,6,11,.......,10086}

                      [  $\because a_{n}=a+(n-1)d$ ]

  and Y= {9,16,23,...........,14128}

           X ∩ Y={16,51,86......}

$\because$ t_n of X∩ Y is less than or equal to 10086

     t_n = 16+(n-1)35 ≤ 10086

$\Rightarrow$    n ≤ 288.7

                    n=288

  $\because$    n(X ∩ Y) =n(X) +n(Y) - n(X ∩ Y )

                      n (X ∩ Y) = 2018+2018-288

                                     =3748