text: The 98 a values form an arithmetic sequence. That is, each successive one is larger than the previous one by a constant amount.\n\nLet the first value be x. (That is, a1 = x.)\nLet the increase amount be i.\nSo a2 = x + i, a3 = x + 2i, etc.\n\nSo the total for the 98 values is 98x + (i + 2i + 3i + ... + 97i).\n(Note that a98 is the 97th term after a1, so it equals a1 plus 97i, not a1 plus 98i.)\n\nWe need a way to simplify the sum of the i's (from 1 i to 97 i's).\nThere is a formula for this. The sum is i (97)(98)/2 = 4753i.\n\nSo now we have:\n98x + 4753i = 2006\n\nThere are lots of values of x and i that will satisfy this equation. In fact, you could choose ANY value for x and then solve for the value of i that will work: i = (2006 - 98x) / 4753\n\nAre x and i supposed to be integers? If so, this problem is of a type called Diophantine equations. These are tricky to solve. \nIn this case, I believe that there is no integer solution, but I'm not positive.