Step-by-Step Solution:

1. Understand the Goal: We need to find the number of elements in the union of two sets, X and Y. The formula for the number of elements in the union of two sets is the Principle of Inclusion-Exclusion: $n(X \cup Y) = n(X) + n(Y) - n(X \cap Y)$

2. Identify Known Values: From the problem statement, we are given:

   • X is the set of the first 2018 terms of an AP. So, n(X) = 2018.
   • Y is the set of the first 2018 terms of another AP. So, n(Y) = 2018. Our main task is to find $n(X \cap Y)$, which represents the number of elements common to both sets X and Y.

3. Define the Arithmetic Progressions:

   • Set X: The AP is 1, 6, 11, ...

     • First term, $a_x = 1$.
     • Common difference, $d_x = 6 - 1 = 5$.
     • The general term (the n-th term) of this AP is $x_n = a_x + (n-1)d_x = 1 + (n-1)5 = 5n - 4$.
     • The terms in X correspond to n = 1, 2, ..., 2018.

   • Set Y: The AP is 9, 16, 23, ...

     • First term, $a_y = 9$.
     • Common difference, $d_y = 16 - 9 = 7$.
     • The general term (the m-th term) of this AP is $y_m = a_y + (m-1)d_y = 9 + (m-1)7 = 7m + 2$.
     • The terms in Y correspond to m = 1, 2, ..., 2018.

4. Find the Common Elements (X ∩ Y): An element is common to both sets if $x_n = y_m$ for some integers n and m within their respective ranges. $5n - 4 = 7m + 2$ $5n - 7m = 6$ This is a linear Diophantine equation. We need to find integer solutions for n and m. Let's express n in terms of m: $5n = 7m + 6$ $n = \frac{7m + 6}{5} = \frac{5m + 2m + 5 + 1}{5} = m + 1 + \frac{2m+1}{5}$ For n to be an integer, (2m+1) must be divisible by 5. Let's test values of m starting from 1:

   • If m=1, 2(1)+1 = 3 (not divisible by 5)
   • If m=2, 2(2)+1 = 5 (divisible by 5) So, m=2 is the smallest possible value for m. This gives $n = \frac{7(2)+6}{5} = \frac{20}{5} = 4$. The first common term corresponds to the 4th term of X ($x_4 = 5(4)-4 = 16$) and the 2nd term of Y ($y_2 = 7(2)+2 = 16$).

5. Generalize the Common Terms: The common terms themselves form an arithmetic progression. The common difference of this new AP is the Least Common Multiple (LCM) of the individual common differences: $d_{common} = LCM(d_x, d_y) = LCM(5, 7) = 35$. The first common term is 16. So the AP of common terms is 16, 51, 86, .... The general form of a common term is $c_k = 16 + (k-1)35$.

6. Count the Number of Common Terms: To find the number of common terms, we need to find how many terms of this common AP fall within the ranges of both sets X and Y.

   • Last term of X: $x_{2018} = 1 + (2018-1)5 = 1 + 2017 \times 5 = 1 + 10085 = 10086$.
   • Last term of Y: $y_{2018} = 9 + (2018-1)7 = 9 + 2017 \times 7 = 9 + 14119 = 14128$. A common term must be less than or equal to the maximum value in both sets. Since all terms in X are less than or equal to 10086, any common term must also satisfy this. $c_k \le 10086$ $16 + (k-1)35 \le 10086$ $(k-1)35 \le 10086 - 16$ $(k-1)35 \le 10070$ $k-1 \le \frac{10070}{35}$ $k-1 \le 287.714...$ $k \le 288.714...$ Since k must be an integer, the maximum value of k is 288. The possible values for k are 1, 2, ..., 288. Thus, there are 288 common elements. So, $n(X \cap Y) = 288$.

7. Calculate n(X U Y): Now we can use the inclusion-exclusion formula: $n(X \cup Y) = n(X) + n(Y) - n(X \cap Y)$ $n(X \cup Y) = 2018 + 2018 - 288$ $n(X \cup Y) = 4036 - 288$ $n(X \cup Y) = 3748$

Thus, the number of elements in the set X $\cup$ Y is 3748.