等差數問題 (急!)

首項是25
等差是-7
設項數是n
n/2[2*25+(n-1)(-7)]=-830
n/2(57-7n)=-830
n(57-7n)=-1660
-7n^2+57n=-1660
7n^2-57n-1660=0
n=20 or n=-11.857
該數列的總項數是20

n (2(25) + (n-1)(-7) = -830 (2)
n (50 + (-7n+7) = -1660
n (50 - 7n + 7) = -1660
50n - 7n^2 + 7n = -1660
57n - 7n^2 = -1660
7n^2 - 57n - 1660 = 0

Sum of AP = (n/2) (2a + (n-1)d) = -830,
where n = no. of term, a = 1st term = 25, d = difference = 18-25 = -7

n (2(25) + (n-1)(-7) = -830 (2)
n (50 + (-7n+7) = -1660
n (50 - 7n + 7) = -1660
50n - 7n^2 + 7n = -1660
57n - 7n^2 = -1660
7n^2 - 57n - 1660 = 0

n = 20

數列的總項數 = 20

Let n be the number of total sum of the series
Let a be the first number of the series
Let d be the difference between two consecutive numbers

From the formula of sum of the series
an + (n(n-1)d)/2 = -830
a = 25, d = -7

25n + n(n-1)(-7)/2 = -830
50n - 7n^2 + 7n = -1660
7n^2 - 57n - 1660 = 0
(7n+83)(n-20) = 0
n = -83/7 (rejected) , n = 20

hope to help you