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?

题目

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?

答案

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