The values of $n$ in question are the multiples of 3 starting with 9. Note that we interpret "distinct" in the problem statement to mean "pairwise distinct" (i.e., no two equal). See the remark below. We first show that such a sequence can only occur when $n$ is divisible by 3. If $d_1$ and $d_2$ are the common differences of the arithmetic progressions $\{x_m, x_{m+1}, x_{m+2}\}$ and $\{x_{m+1}, x_{m+2}, x_{m+3}\}$ for some $m$, then $d_2 \in \{d_1, 2d_1, d_1/2\}$. By scaling we may assume that the smallest common difference that occurs is 1; in this case, all of the common differences are integers. By shifting, we may assume that the $x_i$ are themselves all integers. We now observe that any three consecutive terms in the sequence have pairwise distinct residues modulo 3, forcing $n$ to be divisible by 3. We then observe that for any $m \geq 2$, we obtain a sequence of the desired form of length $3m+3 = (2m-1)+1+(m+1)+2$ by concatenating the arithmetic progressions $$\begin{gather*} (1, 3, \dots, 4m-3, 4m-1), \\ 4m-2, (4m, 4m-4, \dots, 4, 0), 2. \end{gather*}$$ We see that no terms are repeated by noting that the first parenthesized sequence consists of odd numbers; the second sequence consists of multiples of 4; and the remaining numbers $2$ and $4m-2$ are distinct (because $m \geq 2$) but both congruent to 2 mod 4. It remains to show that no such sequence occurs with $n=6$. We may assume without loss of generality that the smallest common difference among the arithmetic progressions is 1 and occurs for $\{x_1, x_2, x_3\}$; by rescaling, shifting, and reversing the sequence as needed, we may assume that $x_1 = 0$ and $(x_2, x_3) \in \{(1,2), (2,1)\}$. We then have $x_4 = 3$ and $$(x_5, x_6) \in \{(4,5), (-1, -5), (-1, 7), (5, 4), (5, 7)\}.$$ In none of these cases does $\{x_5, x_6, 0\}$ form an arithmetic progression.