等差数列中 前n项的和为210，其中前4项的和=40 后4项的和=80 则n=多少？？

设数列为 a1,a2,...,an-1,an
now: a1+a2+a3+a4+an-3+an-2+an-1+an = 40+80 = 120
又有 a1+an = a2+an-1 = a3+an-2 = a4+an-3
so: a1+an = 120/4 = 30
而 前n项和公式 为 210 = （a1+an）n/2
= 30n/2 = 15n
n = 14

(2a1+(n-1)d)*n/2=210
(2a1+3d)*4/2=40
(2a1+(2n-5)d)*4/2=80

2a1=20-3d
2a1=40-(2n-5)d
10=(n-4)d
(20+(n-4)d)*n/2=210
n=210/15=16

14
A1 + A2 + A3 + A4 =40____(1)
An + An-1 +An-2 +An-3=80____(2)
[(1)+(2)]/4 得
A1 + An =30 (A1 + An = A2 + An-1 = A3 +An-2 ......)
210/30=7
n=7*2=14

a1+a2+a3+a4=40
an-3+an-2+an-1+an=80
a1+a2+....+an-1+an=210
设公差为d,

16