已知等差数列{an}.公差d不等于0，a3=4，a2，a5，a11成等比，求an通项公式，

a(n) = a + (n-1)d,n=1,2,...
4 = a(3) = a + 2d,
[a+4d]^2 = [a(5)]^2 = a(2)*a(11) = [a+d][a+10d]
[4+2d]^2 = [4-d][4+8d],
16 + 16d + 4d^2 = 16 + 28d - 8d^2,
0 = 12d^2 - 12d = 12d(d-1),d=1.
a = 4-2d=4-2=2.
a(n)=a+(n-1)d=2+(n-1)=n+1,n=1,2,...

S(n)=2n+n(n-1)/2=n(n+3)/2,
S(5)=5*8/2 = 20,
a(13)=13+1=14<20=S(5).

d=3
A1=-2
Sn=n/2+n*(n-1)*d/2

a(5)^2=a(2)*a(11)
a(5)=a(3)+2d
a(2)=a(3)-d
a(11)=a(3)+8d
把以上三个式子代入a(5)^2=a(2)*a(11)
可得3a(3)d^2-3a(3)d=0
代a(3)=4入上式
则d=0或d=1
因为d不等于0
所以d=1 a(1)=2
所以a(13）=a(3)+10d=14
S5=(2+6)*5/2=20
所以S5>a13