已知等差数列{an}的公差d>0,其前n项和为Sn,若S3=12,且2a1,a2,1+a3成等比数列.（1）求{an}的

a2^2=2a1*(a3+1)
(a1+d)^2=2a1*(a1+d+1)
a1^2+2a1d+d^2=2a1^2+2a1d+2a1
a1^2+2a1-d^2=0
s3=3a1+3*2*d/2=3a1+3d=12
a1+d=4
d=4-a1代入上式得
a1^2+2a1-(4-a1)^2=0
a1^2+2a1-16+8a1-a1^2=0
10a1=16
a1=1.6
d=2.4
所以
an=a1+(n-1)*d
=1.6+2.4n-2.4
=2.4n-0.8