题干

设数列{an}的前n项和为Sn,满足(1﹣q)Sn+qan=1,且q(q﹣1)≠0.

(Ⅰ)求{an}的通项公式;

(Ⅱ)若S3,S9,S6成等差数列,求证:a2,a8,a5成等差数列.

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答案(点此获取答案解析)

解:(Ⅰ)当n=1时,由(1﹣q)S1+qa1=1,a1=1.

当n≥2时,由(1﹣q)Sn+qan=1,得(1﹣q)Sn1+qan1=1,两式相减得:(1﹣q)an+qan﹣qa

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