數學問題～幫我解決．

若等差級數25+18+11+4...的總和為-830,求該數列的總項數。

T(1) = 25

等差=18 - 25= -7

設 n 是該數列的總項數

n[2(25) + (n-1)(-7)]/2 = -830

n(50 -7n + 7) = -1660

57n -7n2 = -1660

7n2 - 57n - 1660 = 0

(7n+83)(n-20) = 0

7n+83= 0 or n-20=0

n = -83/7 (捨去) or n = 20

該數列的總項數是20項

(n) (Item) (Sum of accumulated item)
12525
21843
31154
4458
5-355
6-1045
7-1728
8-244
9-31-27
10-38-65
11-45-110
12-52-162
13-59-221
14-66-287
15-73-360
16-80-440
17-87-527
18-94-621
19-101-722
20-108-830
21-115-945
22-122-1067
23-129-1196
24-136-1332

I think n=20, I used EXCEL to calculate.