题目内容
15.设等比数列
15.-2
解析:∵S n+1,Sn,S n+2成等差数列,
∴Sn-S n+1=S n+2-Sn,即-a n+1=a n+2+a n+1.
∴an+2=-2an+1.∴q=
=-2.
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题目内容
15.设等比数列
15.-2
解析:∵S n+1,Sn,S n+2成等差数列,
∴Sn-S n+1=S n+2-Sn,即-a n+1=a n+2+a n+1.
∴an+2=-2an+1.∴q=
=-2.