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

Here $X = {1, 6, 11, \dots, 10086}$

and $Y = {9, 16, 23, \dots, 14128}$

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, $k^{th}$ term = $16 + (k - 1) 35 \leq 10086$

i.e. $k \leq \frac{10105}{35} ∴ k \leq 288$ (as $k$ is to be an integer)

Hence, $\begin{aligned} n (X \cup Y) = n (X) + n (Y - n (X \cap Y) \\ = 2018 + 2018 - 288 = 3748 \end{aligned}$

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 $X \cup Y$ is