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 

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

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

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