The sum of four positive integers that form an arithmetic sequence is 46. Of all such possible sequences, what is the greatest possible third term?
If the first term is a, and the common difference is d, then the second, third, and fourth terms are a + d, a + 2d, and a + 3d, respectively. The sum of these four terms is 46, so I have the equation a + (a + d) + (a + 2d) + (a + 3d) = 46. Simplifying, I get 4a + 6d = 46, or 2a + 3d = 23. I want to find the maximum value of a + 2d, which is the third term. Since a and d are positive integers, I can try different values of d and see how they affect a and a + 2d. If d = 1, then 2a + 3 = 23, so a = 10 and a + 2d = 12. If d = 2, then 2a + 6 = 23, so a = 8.5, which is not an integer. If d = 3, then 2a + 9 = 23, so a = 7 and a + 2d = 13. If d = 4, then 2a + 12 = 23, so a = 5.5, which is not an integer. If d = 5, then 2a + 15 = 23, so a = 4 and a + 2d = 14. If d = 6, then 2a + 18 = 23, so a = 2.5, which is not an integer. If d = 7, then 2a + 21 = 23, so a = 1 and a + 2d = 15. If d > 7, then 2a + 3d > 23, so there are no more positive integer solutions. Therefore, the maximum possible value of a + 2d is 15, when d = 7 and a = 1. The corresponding arithmetic sequence is 1, 8, 15, 22, and the third term is 15. 