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  4. 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 ∪ Y is .

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Q. 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 ∪ Y is _____.

JEE AdvancedJEE Advanced 2018

Solution:

$X: 1,6,11, \ldots 10086$
$Y: 9,16,23, \ldots, 14128$
$X \cap Y: 16,51,86, \ldots$
Let $m=n(X \cap Y)$
$16+(m-1) \times 35 \leq 10086 $
$ m \leq 288.71 \,m=288$
$n(X \cup Y)=n(X)+n(Y)-n(X \cap Y)$
$=2018+2018=288=3748$