19等差級數 (2+1/2)+(11+1/4) https://www.flickr.com/photos/101292575@N06/20627736824/in/dateposted-public/?

S(n) = (12又1/2) + (11又1/4) + 10 + (8又3/4) + ......
首項 = 12又1/2 = 25/2
公差 = (11又1/4) - (12又1/2) = -5/4

S(n) = n [2a + (n - 1)d] / 2
n [2×(25/2) + (n - 1) × (-5/4)] / 2 = 61又1/4
n [25 + (-5n + 1) / 4] /2 = (245/4)
n [25 + (-5n + 5) / 4] × 4 = (245/2) × 4
n [100 + (-5n + 5)] = 490
-5n² + 105n = 490
n² - 21n + 98 = 0
(n - 7)(n - 14) = 0
n = 7 或 n = 14 ...... 答案選(C)