Question

已知Sn是首项唯a的等比数列{an}的前n项的和，S3,S9,S6成等差数列。（1）求证：a2,a8,a5成等差数列

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Answer 1

(1)若q=1时
S3+S6=2S9
3a+6a=2*9a不成立
(2)若q≠1
S3+S6=2S9
a(1-q^3)/(1-q)+a(1-q^6)/(1-q)=2a(1-q^9)/(1-q)
1+(1+q^3)=2(1+q^3+q^6)
2q^6+q^3=0
q=0或q^3=-1/2
a8=a2*q^6=1/4a2
a5=a2*q^3=-1/2a2
2a8=a2+a5
a2,a8,a5成等差数列