Question

Let $X$ be the set consisting of the first $2018$ terms of the arithmetic progression $1,6,11,\cdots ,$ and $Y$ be the set consisting of the first $2018$ terms of the arithmetic progression $9,16,23,\cdots$ . Then, the number of elements in the set $X\cup Y$ is

Answer 1

$X=\{1,6,11,16,\cdots \}{2018}^{th}\text{term}=1+2017\times 5=10086X=\{1,6,11,16,\cdots ,10086\}$

$Y=\{9,16,23,30,\cdots \}{2018}^{th}\text{term}=9+2017\times 7=14128Y=\{9,16,23,\cdots ,14128\}$

$X\cap Y=\{16,51,86,\cdots \}$
The last digit in $X\cap Y\le 10086$
$\therefore 16+35(m-1)\le 10086\Rightarrow m\le 288.7$
As $m$ is an integer so,
Number of terms,$n(X\cap Y)=288$
Number of elements,
$n(X\cup Y)=n\left(X\right)+n\left(Y\right)-n(X\cap Y)=2018+2018-288=3748$