等差数列｛an｝的前n项和为Sn,设S3=12,且2a1,a2,a3 1成等比数列,求Sn?

S3=a1+a2+a3=a1+(a1+d)+(a1+2d)=3a1+3d=12
所以a1+d=4 (1)
2a1,a2,a3+1成等比数列
则a2²=2a1*(a3+1)
即(a1+d)²=2a1(a1+2d+1)
(1)代入4²=2a1*(4+d+1)
a1*(5+d)=8 a1(9-a1)=8
a1²-9a1+8=0
解得a1=1或8
代入(1) d=3或-4
所以an=1+3(n-1)=3n-2
或an=8-4(n-1)=12-4n

S3=a1+a2+a3=3a2=12
a2=4
设公差为d,则a1=a2-d=4-d a3=a2+d=4+d
2a1、a2、a3+1成等比数列,则
a2²=(2a1)(a3+1)
2(4-d)(4+d+1)=16
整理,得
(d-4)(d+5)+8=0
d²+d-12=0
(d+4)(d-3)=0
d=-4或d=3
d=-4时,a1=a2-d=4-(-4)=8 Sn=na1+n(n-1)d/2=8n-4n(n-1)/2=10n-2n²
d=3时,a1=a2-d=4-3=1 Sn=na1+n(n-1)d/2=n+3n(n-1)/2=n(3n-1)/2