The positive integer $k$ and the set $A$ of distinct integers from $1$ to $3k$ inclusively are such that there are no distinct $a$, $b$, $c$ in $A$ satisfying $2b = a + c$. The numbers from $A$ in the interval $[1, k]$ will be called [i]small[/i]; those in $[k + 1, 2k]$ - [i]medium[/i] and those in $[2k + 1, 3k]$ - [i]large[/i]. It is always true that there are [b]no[/b] positive integers $x$ and $d$ such that if $x$, $x + d$, and $x + 2d$ are divided by $3k$ then the remainders belong to $A$ and those of $x$ and $x + d$ are different and are:
a) small? $\hspace{1.5px}$  b) medium? $\hspace{1.5px}$  c) large?
([i]In this problem we assume that if a multiple of $3k$ is divided by $3k$ then the remainder is $3k$ rather than $0$[/i].)
1. **Part (a): Small Numbers**

   We need to determine if there exist positive integers \( x \) and \( d \) such that the remainders of \( x \), \( x+d \), and \( x+2d \) when divided by \( 3k \) are in \( A \), and the remainders of \( x \) and \( x+d \) are both small.

   Let \( k = 3 \) and \( A = \{1, 2, 9\} \). Choose \( x = 2 \) and \( d = 8 \).

   - The remainder of \( x \) when divided by \( 3k \) is \( 2 \).
   - The remainder of \( x + d = 2 + 8 = 10 \) when divided by \( 3k \) is \( 10 \mod 9 = 1 \).
   - The remainder of \( x + 2d = 2 + 2 \cdot 8 = 18 \) when divided by \( 3k \) is \( 18 \mod 9 = 0 \), but since we consider \( 3k \) as the remainder, it is \( 9 \).

   Since \( 2 \) and \( 1 \) are both small (in the interval \([1, k]\)), the answer is \(\boxed{\text{no}}\).

2. **Part (b): Medium Numbers**

   We need to show that there are no positive integers \( x \) and \( d \) such that the remainders of \( x \), \( x+d \), and \( x+2d \) when divided by \( 3k \) are in \( A \), and the remainders of \( x \) and \( x+d \) are both medium.

   Assume the answer is no, so there exist integers \( 0 < x \le 3k \) and \( 0 < d < 3k \) that satisfy the conditions. We consider two cases:

   **Case 1:** \( x + d \le 3k \)

   - Since the remainders of \( x \) and \( x + d \) when divided by \( 3k \) are medium, \( x, x + d \in [k+1, 2k] \) and \( d < k \).
   - Hence, \( x + 2d = (x + d) + d < 2k + k = 3k \).
   - So, \( x, x + d, x + 2d \in A \) and \( 2(x + d) = x + (x + 2d) \), which is a contradiction.

   **Case 2:** \( x + d > 3k \)

   - Since the remainder of \( x + d \) when divided by \( 3k \) is medium, \( x + d \in [4k + 1, 5k] \) and the remainder when \( x + d \) is divided by \( 3k \) is \( x + d - 3k \in A \).
   - Since \( x \in [k+1, 2k] \), then \( d = (x + d) - x \ge 4k + 1 - 2k = 2k + 1 > 2k \).
   - Hence, \( 6k = 4k + 2k < (x + d) + d < 5k + 3k = 8k \), and the remainder when \( x + 2d \) is divided by \( 3k \) is \( x + 2d - 6k \in A \).
   - Obviously, \( x \in A \). So \( x, x + d - 3k, x + 2d - 6k \in A \) and \( 2(x + d - 3k) = x + (x + 2d - 6k) \), which is a contradiction.

   Therefore, the answer to (b) is \(\boxed{\text{yes}}\).

3. **Part (c): Large Numbers**

   We need to determine if there exist positive integers \( x \) and \( d \) such that the remainders of \( x \), \( x+d \), and \( x+2d \) when divided by \( 3k \) are in \( A \), and the remainders of \( x \) and \( x+d \) are both large.

   Let \( k = 3 \) and \( A = \{1, 8, 9\} \). Choose \( x = 8 \) and \( d = 1 \).

   - The remainder of \( x \) when divided by \( 3k \) is \( 8 \).
   - The remainder of \( x + d = 8 + 1 = 9 \) when divided by \( 3k \) is \( 9 \).
   - The remainder of \( x + 2d = 8 + 2 \cdot 1 = 10 \) when divided by \( 3k \) is \( 10 \mod 9 = 1 \).

   Since \( 8 \) and \( 9 \) are both large (in the interval \([2k + 1, 3k]\)), the answer is \(\boxed{\text{no}}\).