已知Sn是等比数列〔Aa〕的前n项和．S3.S9.S6成等差数列．求证A2.A8.A5成等差数列

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已知：Sn是等比数列{An]的前n项和,S3、S9、S6成等差数列；
求证：A2、A8、A5成等差数列.
证明：由已知设An=A1q^(n-1),q为公比且不为0.
则q不为0也不为1时,Sn=[(q^n-1)/(q-1)]A1；
当q=1时,Sn=nA1；
∵2S9=S3+S6,
∴2[(q^9-1)/(q-1)]A1=[(q^3-1)/(q-1)]A1+[(q^6-1)/(q-1)]A1,
∴2q^9=q³+q^6,∴q³(2q³+1)(q³-1)=0,
∴q³=1或-1/2,∴q=1或(-1/2)^(1/3),
当q=1时,An=A1,则2A8-(A8+A5)=2A1-(A1+A1)=0,得证；
当q=(-1/2)^(1/3)时,An=A1[(-1/2)^(1/3)]^(n-1)；
2A8-(A8+A5)
=2×A1[(-1/2)^(1/3)]^(8-1)
-{[A1[(-1/2)^(1/3)]^(2-1)]+[A1[(-1/2)^(1/3)]^(5-1)]}
=(A1/2)(-1/2)^(1/3)-[A1(-1/2)^(1/3)+(-A1/2)(-1/2)^(1/3)]
=(A1/2)(-1/2)^(1/3)-(A1/2)(-1/2)^(1/3)
=0,得证.