设数列{an}的前n项和为Sn,其中an不等于0.a1为常数,且-a1,Sn,a(n+1)成等差数列

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设数列{an}的前n项和为Sn,其中an不等于0.a1为常数,且-a1,Sn,a(n+1)成等差数列
设BN=1-SN,是否存在A1,使数列{BN}为等差数列?若存在,求出A1,若不存在说明理由

由三者成等差数列知a(n+1)-Sn=Sn+a1,即2*Sn=a(n+1)-a1
由bn=1-Sn知,b(n+1)-bn=1-S(n+1)-1+Sn=Sn-S(n+1)=0.5*[a(n+1)-a1-a(n+2)+a1]=0.5*[a(n+1)-a(n+2)]很明显如果数列{an}是等差数列,则{bn}就是等差数列.
若{an}为等差数列则2*Sn=a1+a(n)=a(n+1)-a1可推出,2a1=a(n+1)-a(n)恒成立,故不论a1取什么,均为等差数列

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