Solution
Alex Y
I'll take this step-by-step, in case anyone is still working on it and wants to only read far enough to get the needed direction.
Each player's scores being an arithmetic progression means that we can express the match winner's scores in the three games as A, A+a, A+2a; and the loser's as B, B+b, B+2b.
A's margin of victory in the three games (negative in the game he lost) can be expressed as M, M+m, M+2m, where M=A-B and m=a-b.
We know that the winner of the match must have won the last game, so M+2m>0. But we also know that the winner of the match lost one game, and since the margins of victory form an arithmetic progression, that must have been the first game.
Now we use the fact that a game must be won by at least two points to say from his loss of the first game that M <= -2 and from his victory in the second game that M+m >= 2.
Hence m >= 4, and M+2m >= 6.
Since the match winner won the last game by at least six points, that means his score in the last game was 21.
Since we are told that no-one's score was the same in all three games, and we know that the winner must have gotten at least 21 in another game to win it, that means that the match winner's score in the first two games were greater than 21.
The only way anyone scores more than 21 points in a game is for that game to end in a 2-point margin of victory. Thus, we know that M = -2 and M+m = 2, so M+2m = 6.
Therefore, the score in the last game was 21-15.
Interestingly, we have no way of knowing the scores of the first two games. The match may have been 23-25, 22-20, 21-15, or just as easily (if less probably) been 41-43, 31-29, 21-15.