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 \cup Y is: Sequence & Series (Mathematics)

Here $X=\{1,6,11, \ldots, 10086\}$ and $Y=\{9,16,23, \ldots, 14128\}\\$ The intersection of $X$ and $Y$ is an A.P. with 16 as first term and $\\$ 35 as common difference. $\\$ The series becomes $16,51,86, \ldots. \\$ Now, $k^{\text {th }}$ term $=16+(k-1) 35 \leq 10086$ i.e. $k \leq \frac{10105}{35} \\ \therefore k \leq 288$ (as $k$ is to be an integer) $\\$ Hence, $n(X \cup Y)=n(X)+n(Y)-n(X \cap Y)$ =2018+2018-288=3748$

Explanation

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