已知Sn是等比数列｛An｝的前n项和,S3,S9,S3成等差数列,求证a2,a8,a5成等差数列

题目应该为S3,S9,S6成等差数列 证明：首先根据等差数列的通项公式an=a1+（n-1）d 则a(n+1)=a1+（n)d a（n+2)=a1+（n+1）d 很显然等差数列有an+a(n+2)=2a(n+1) 根据等比数列的求和公式：Sn=a1/(1-q)-a1/(1-q)*q^n 则S...