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
这题说来话长.不说了好
人气:378 ℃ 时间:2020-05-01 21:20:02
解答
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...
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