已知SN是等比数列{AN}的前N项和,S3,S9,S6成等差数列,求证A2,A8,A5成等差数列

2025-06-28 13:48:52
推荐回答(1个)
回答1:

s3=a1(1-q^3)/(1-q)
s9=a1(1-q^9)/(1-q)
s6=a1(1-q^6)/(1-q)
s6-s9=s9-s3,代入得:2q^9=q^3+q^6
即:2q^7=q+q^4
若a2,a8,a5成等差数列
那么a5-a8=a8-a2,a2+a5=2a8
即:a1q+a1q^4=2a1q^7
因为2q^7=q+q^4
所以a2,a8,a5成等差数列