Question

等差｛an｝前n项和为sn，设s3＝12，且2A1，A2，A3+1成等比数列，求An.Sn

Answer 1

由等差数列性质，S[3]=a[1]+a[2]+a[3]=3a[2]=12
所以a[2]=4
设等差数列公差为d，那么a[1]=4-d,a[3]=4+d
2a[1] a[2] a[3]+1等比，即a[2]²=2a[1]*(a[3]+1)
即16=2(4-d)(5+d)
解得d=3或-4
当d=3时，a[1]=4-3=1,
a[n]=1+3(n-1)=3n-2
S[n]=3*n(n+1)/2 -2n=(1/2)n(3n-1)
当d=-4时，a[1]=8
a[n]=8-4(n-1)=12-4n
S[n]=12n-4n(n+1)/2=2n(5-n)

Answer 2

S3=a1+a2+a3=3a2=12
a2=4
数列成等比数列 a2^2=2a1(a3+1)
a2^2=2(a2-d)(a2+d+1)
a2=4代入，整理，得
16+2(d-4)(d+5)=0
d^2+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=-2n^2+10n
d=3时，a1=a2-d=4-3=1 Sn=na1+n(n-1)d/2=n+3n(n-1)/2=3n^2/2 -n/2

Answer 3

S3=3a1+3d=12，a1+d=4
a2=a1+d=4，a3=a1+2d=d+4
2a1(a3+1)=2a1(d+5)=a2^2=16
a1(d+5)=8
(4-d)(d+5)=8
d^2+d-12=0
d=-4或d=3
a1=8或a1=1
an=-4n+12或an=3n-2
Sn=-2n^2+10n或Sn=(3/2)n^2-(1/2)n
n为正整数。