Question

已知等比数列{an}的前n项和为Sn，a1=1，且满足Sn，Sn+2，Sn+1成等差数列，则a3=

Answer 1

2S(n+2)=Sn+S(n+1)
则2S3=S1+S2
2(a1+a2+a3)=a1+(a1+a2)
所以a2=-2a3
q=a3/a2=-1/2
所以a3=a1*q²=1/4

Answer 2

C ∵Sn、Sn+2、Sn+1成等差数列,∴Sn+2-Sn=Sn+1-Sn+2.∴an+2+an+1=-an+2,∴an+2/an+1=</B>.1/2 又a1=1,∴a3=1/4.