text: how would one go about solving the arithmetic sequence of a1+a2+a3...+a98=2006?   the 98 a values form an arithmetic sequence. that is, each successive one is larger than the previous one by a constant amount.  let the first value be x. (that is, a1 = x.) let the increase amount be i. so a2 = x + i, a3 = x + 2i, etc.  so the total for the 98 values is 98x + (i + 2i + 3i + ... + 97i). (note that a98 is the 97th term after a1, so it equals a1 plus 97i, not a1 plus 98i.)  we need a way to simplify the sum of the i's (from 1 i to 97 i's). there is a formula for this. the sum is i (97)(98)/2 = 4753i.  so now we have: 98x + 4753i = 2006  there 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  are x and i supposed to be integers? if so, this problem is of a type called diophantine equations. these are tricky to solve.  in this case, i believe that there is no integer solution, but i'm not positive.