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 $X\cup Y$ is

Let  $m^{\text{th }}$ term of first A.P be same as the ${11}^{th}$ term of the second A.P. Then,

$\begin{array}{l}1+5(m-1)=9+7(n-1)\Rightarrow 5m-4=7n+2\\ \Rightarrow 5(m+3)=7(n+3)\Rightarrow \frac{m+3}{5}=\frac{n+3}{7}=r\text{ (say) }\end{array}$

But, $m\le 2018,n\le 2018$

$\begin{array}{l}\Rightarrow 5r-3\le 2018,7r-3\le 2018\\ \Rightarrow r\le 404.2\text{ and }r\le 288.7\\ \Rightarrow r=288\end{array}$

Thus, there are $288$ common terms.

$n\left(X\right)=2018,n\left(Y\right)=2018$ and $n(X\cap Y)=288$

Hence,

$n(X\cup Y)=n\left(X\right)+n\left(Y\right)-n(X\cap Y)=2018+2018-288=3748$