等差数列中a1,a4,a5成等比数列，则a4^2=a1*a5,(a1+3d)^2=a1*(a1+4d),展开a1^2+6a1*d+9d^2=a1^2+4a1*d，解的a1=-9/2d,将a1用-9/2d代换，所以a1=-9/2d,a4=-3/2d,a5=-1/2d,,此时公比为1/3，一种特殊情况就是等差数列的所有项都相等，即公差为0，那么a1=a4=a5,即公比为1，原题得证
