Let $X$ be the set consisting of the first 2018 terms of the arithmetic progression $1,6,11, \ldots$

Question

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

Professor PC Thomas Classes · Answer

(3748) : 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, $k^{	ext {th }}$ term $=16+(k-1) 35 \leq 10086$
i.e. $k \leq \frac{10105}{35}$
$	herefore \quad k \leq 288$ (if $k$ is to be an integer)
Hence, $(X \cup Y)=n(X)+n(Y)-n(X \cap Y)$ $=2018+2018-288=3748$