7-3 = 4, 11=7 = 4, 15 - 11 = 4, so common difference -= d = 4 Sn =[2a1 + (n-1)d] where a1 = first term = 3. Sn = [2a1 + (n-1)d] < 500 [2(3) + (n-1)(4)] < 500 Multiply both sides by 2 to get rid of the fraction: n[2(3) + (n-1)(4)] < 1000 n[6 + 4(n-1)] < 1000 n[6 + 4n - 4] < 1000 n[2 + 4n] < 1000 2n + 4nē < 1000 Divide through by 2 n + 2nē < 500 Get 0 on the right and arrange terms on the left in descending order: 2nē + n - 500 < 0 Using the quadratic formula we see that this has critical values approximately 15.6 and -16.1 We ignore the negative critical value and take the largest integer that does not exceed 15.1, which is 15. So the first 15 terms have a sum less than 500. answer = 15 Checking: 3+7+11+15+19+23+27+31+35+39+43+47+51+55+59 = 465 The 16th term would be 63, which would give a sum 528, which is over 500, so answer = 15 is true. Edwin