Question

已知等差数列An的前n项和为Sn=120，且S4=20，Sn-4=60，则n=?

Answer 1

S4 = 4a1 + 4(4-1)d/2 = 4a1 + 6d = 20
2a1 + 3d = 10
2a1 = 10 - 3d (1)
Sn = na1 + n(n-1)d/2 = 120
2na1 + n(n-1)d = 240 (2)
n(10-3d) + n(n-1)d = 240
10n -3nd + (n-1)nd = 240 (3)
Sn - Sn-4 = an + an-1 + an-2 + an-3 = a1 + (n-1)d + a1 + (n-2)d + a1 + (n-3)d + a1 + (n-4)d
= 4a1 + (4n-10)d
= 2(10-3d) + (4n-10)d
= 20 -6d +4nd -10d = 4nd -16d +20 = 120 - 60 = 60
4nd -16d = 40
nd - 4d = 10
nd = 10 + 4d (4)
(4)带入(3):
10n -3(10+4d) +(n-1)(10+4d) = 240
10n -30 -12d + 10n -10 + 4nd -4d = 240
20n - 16d + 4nd = 280
5n -4d +nd = 70
5n -4d + 10 + 4d = 70
5n = 60
n = 12
(d = 5/4, a1 = 25/8)