设等比数列(an},若S3+S6=2S9,求证:a2,a8,a5成等差数列

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/07/19 15:29:07

证：
S3+S6=2S9
a1(q^3-1)/(q-1)+a1(q^6-1)/(q-1)=2a1(q^9-1)/(q-1)
q^3-1+q^6-1=2q^9-2
q^3+q^6=2q^9
2q^6-q^3-1=0
(q^3-1)(2q^3+1)=0
q^3=1或q^3=-1/2
q^3=1时
a2+a5=a1q+a1q^4=a1q(1+q^3)=2a1q
2a8=2a1q^7=2a1q
2a8=a2+a5
a2,a8,a5成等差数列.
q^3=-1/2时
a2+a5=a1q+a1q^4=a1q(1+q^3)=a1q/2
2a8=2a1q^7=2a1q(q^6)=a1q/2
2a8=a2+a5
a2,a8,a5成等差数列.
综上,得a2,a8,a5成等差数列

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