<正>性质1在等比数列{an}中,相邻两项之比相等.推论1在等比数列等.{an}中,序号(或项数)间隔相等的两项之比相即:ar+m/ar=ak+m/ak=qm(q为公比).证明设等比数列n{an}的首项为a1,公比为q,则有an=a1q(n-1,)所以ar+m/ar=a1·qr+m-1/a1·qr-1=qm,同理可得ak+m/ak=qm,所以ar+m/ar=ak+m=ak=qm(q为公比).