等差数列an的前n项和为Sn,设S3=122,且2a1.a2.,a3+1成等比数列,求Sn

S3应该是等于12吧
{an}是等差数列
S3=a1+a2+a3=3a2=12
a2=4
设公差为d
a1=4-d a3=4+d
2a1,a2,a3+1成等比数列
(a2)^2=2a1·(a3+1)
4^2=2(4-d)(4+d+1)
8=(4-d)(d+5)
8=20-d-d^2
d^2+d-12=0
(d+4)(d-3)=0
d=-4 或 d=3
若d=-4,则a1=8,an=a1+d(n-1)=8-4(n-1)=12-4n
Sn=(a1+an)n/2=(8+12-4n)n/2=-2n^2+10n
若d=3,则a1=1,an=a1+d(n-1)=1+3(n-1)=3n-2
Sn=(a1+an)n/2=(1+3n-2)n/2=(3/2)n^2-(1/2)n