The sequence is $2, 4, 6, \ldots, 2n-2$ .
Proof 1
We will prove that any sequence $x_1, \ldots, x_{n-1}$ , that satisfies
the given conditions, is an
arithmetic progression with $x_1$ as both the first term and the
increment. Once this is proved, condition (b) implies that $x_1 + x_{n-1} = x_1 + (n-1)x_1 = nx_1 = 2n$ . Therefore $x_1 = 2$ ,
and the sequence is just the even numbers from $2$ to $2n-2$ . The
sequence of successive even numbers clearly satisfies all three conditions,
and we are done.
First a degenerate case.
If $n = 2$ , there is only one element $x_1$ , and condition (b) gives $x_1 + x_1 = 4$ or $x_1 = 2$ . Conditions (a) and (c) are vacuously
true.
Otherwise, for $n > 2$ , we will prove by induction on $m$ that the
difference $x_{n-m} - x_{n-1-m} = x_1$ for all $m \in [1, n-2]$ ,
which makes all the differences $x_{n-1} - x_{n-2} = \ldots = x_2 - x_1 = x_1$ , i.e. the sequence is an arithmetic progression with $x_1$ as the first term and increment as promised.
So first the $m=1$ case. With $n > 2$ , $x_{n-2}$ exists and is less
than $x_{n-1}$ by condition (a). Now since by condition (b) $x_1 + x_{n-1} = 2n$ , we conclude that $x_1 + x_{n-2} < 2n$ , and therefore
by condition (c) $x_1 + x_{n-2} = x_k$ for some $k$ . Now, since $x_1 > 0$ , $x_k > x_{n-2}$ and can only be $x_{n-1}$ . So $x_1 + x_{n-2} = x_{n-1}$ .
Now for the induction step on all values of $m$ .
Suppose we have shown that for all $i \le m$ , $x_1 + x_{n-1-i} = x_{n-i}$ . If $m = n-2$ we are done, otherwise $m < n-2$ , and by
condition (c) $x_1 + x_{n-2-m} = x_k$ for some $k$ . This $x_k$ is
larger than $x_{n-2-m}$ , but smaller than $x_1 + x_{n-1-m} = x_{n-m}$ by the inductive hypothesis. It then follows that $x_1 + x_{n-2-m} = x_{n-1-m}$ , the only element of the sequence between $x_{n-2-m}$ and $x_{n-m}$ . This establishes the result for $i=m+1$ .
So, by induction $x_1 + x_{n-1-m} = x_{n-m}$ for all $m \in [1, n-2]$ ,
which completes the proof.
Proof 2
Let $S=\{x_1,x_2,...,x_{n-1}\}$ .
Notice that $$x_1<x_1+x_1<x_1+x_2<\dots <x_1+x_{n-2}<2n.$$ Then by condition (c), we must have $x_1,x_1+x_1,...,x_1+x_{n-2}\in S$ . This implies that $x_1=x_1,x_1+x_1=x_2,...,x_1+x_{n-2}=x_{n-1}$ , or that $x_k=kx_1$ . Then we have $x_1+x_{n-1}=n(x_1)=2n\rightarrow x_1=2$ , and the rest is trivial.