设等差数列{an}的首项a1为a,前n项和为Sn

若Sn,S(n+1),S(n+2)构成等比数列
[S(n+1)]^2=Sn*S(n+2)
=(S(n+1)-a(n+1))*(S(n+1)+a(n+2))=[S(n+1)]^2+a(n+2)*S(n+1)-a(n+1)*S(n+1)-a(n+2)*a(n+1)
得a(n+2)*S(n+1)=a(n+1)*S(n+1)+a(n+2)*a(n+1)=a(n+1)[S(n+1)+a(n+2)]=a(n+1)*S(n+2)
得a(n+2)/a(n+1)=S(n+2)/S(n+1)
得a(n+1)/an=S(n+1)/Sn
又若Sn,S(n+1),S(n+2)构成等比数列
所以a(n+2)/a(n+1)=S(n+2)/S(n+1)=S(n+1)/Sn=a(n+1)/an
即{an}又为等比数列
所以an=a(既等差又等比)
Sn=na
S(n+1)=(n+1)a
S(n+2)=(n+2)a
再由[S(n+1)]^2=Sn*S(n+2)
得[(n+1)a]^2=(na)*(n+2)a
解出a=0
得Sn=S(n+1)=S(n+2)=0
显然不是等比数列
故导出矛盾,原假设不成立.
故Sn,S（n+1）,S(n+2)不构成等比数列

2不是总成立，如果等差的公差是0，它构成等比
当d不为时，S(n+1) = S(n)+d, S(n+2) = S(n)+2d
S(n)S(n+2) = S(n)^2 + 2dS(n)
[S(n+1)]^2 = S(n)^2 + 2dS(n) + d^2
显然上式只有d=0时才成立，也就是说只有d=0时他们才能成为等比数列