设等比数列公比q
Sn=1*(1-q^n)/(1-q)
若数列{Sn}为等差数列'
2Sn=Sn-1+Sn+1
2*(1-q^n)/(1-q)=1*(1-q^n-1)/(1-q)+1*(1-q^n+-)/(1-q)
得2q^n=q^n-1+q^n+1
2q=1+q^2
得q=1
则an=1
Sn=n
故数列{Sn}为等差数列 则它的公差为1