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(B) The $n$-th term of $X$ is $a_n = 1 + (n-1)5 = 5n - 4$. For $n=2018$,$a_{2018} = 5(2018) - 4 = 10086$.The $n$-th term of $Y$ is $b_n = 9 + (n-1)7 =

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$3748$

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(B) The $n$-th term of $X$ is $a_n = 1 + (n-1)5 = 5n - 4$. For $n=2018$,$a_{2018} = 5(2018) - 4 = 10086$.The $n$-th term of $Y$ is $b_n = 9 + (n-1)7 = 7n + 2$. For $n=2018$,$b_{2018} = 7(2018) + 2 = 14128$.Common terms $X \cap Y$ satisfy $5n - 4 = 7m + 2$,which implies $5n = 7m + 6$. The smallest solution is $n=4, m=2$,giving $16$. The common difference is $	ext{lcm}(5, 7) = 35$.The common terms are $16, 51, 86, \dots$. The general term is $c_k = 16 + (k-1)35 = 35k - 19$.We need $c_k \leq 10086$ (since $c_k$ must be in $X$ and $Y$): $35k - 19 \leq 10086 \implies 35k \leq 10105 \implies k \leq 288.71$.Thus,$n(X \cap Y) = 288$.Using the inclusion-exclusion principle: $n(X \cup Y) = n(X) + n(Y) - n(X \cap Y) = 2018 + 2018 - 288 = 3748$.

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

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