Let X be the set consisting of the first 2023 terms of arithmetic progression 2, 7, 12, … and Y be the set consisting of the first 2023 terms of arithmetic progression 3, 10, 17, 24, …… Then number of elements in the set $X \cap Y$ is …….

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answer is 289.

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Detailed Solution

X : 2, 7, 12, …..…., 10087, 10092
Y : 3, 10, 17, ……., 14122, 14129
$X \cap Y$ :17, 52, 87, ………..
Let $m = n (X \cap Y)$
$∴ 17 + (m - 1) \times 35 \leq 10112$
$\Rightarrow m = 289$

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