Question

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

Answer 1

解：因为S3=12
所以a1+a2+a3=3a2=12
故a2=4
又2a1，a2，a3+1成等比数列
所以(a2)^2=2a1*(a3+1)
即4^2=2*(a2-d)*(a2+d+1)=2*(4-d)*(5+d)
即d^2-d-12=0
所以d=3或d=-4
当d=3时
a1=a2-d=1
Sn=na1+n(n-1)d/2=n+3n(n-1)/2=n(3n-1)/2

当d=-4时
a1=a2-d=8
Sn=na1+n(n-1)d/2=8n-4n(n-1)/2=10n-2n^2

Answer 2

等比则a2²=2a1(a3+1)
(a1+d)²=2a1(a1+2d+1)
d²=a1²+2a1d+2a1

S3=a1+a2+a3=3a1+3d=12
a1=4-d
代入
d²+d-12=0
d=-4,d=3
a1=8,a1=1
则a1=8,an=-4n+12
a1=1,an=3n-2

Sn=-2n²+10n或Sn=(3n²-3n)/2