已知Sn是等比数列｛an}的前n项和,a2.a8,a5成等差数列

题目

已知Sn是等比数列｛an}的前n项和,a2.a8,a5成等差数列
（1）求等比数列｛an}的公比q
得出q=1或q=-1/2的根号三次方
（2）判断s3,s9,s6是否成等差数列
①q=1时不成等差
②q=-1/2的根号三次方 时.

答案

a2=a1q
a8=a1q^7
a5=a1q^4
2a8=a2+a5
2a1q^7=a1q+a1q^4
2q^6=1+q^3
2q^6=1+q^3
2q^6-q^3-1=0
(2q^3+1)(q^3-1)=0
q^3=-1/2 或q^3=1
q=-1/2^1/3 或q=1
s3=a1(1-q^3)/(1-q)
s9=a1(1-q^9)/(1-q)
s6=a1(1-q^6)/(1-q)
s3+s6=a1(1-q^3)/(1-q)+a1(1-q^6)/(1-q) 将q^3=-1/2 代入
=a1(1+1/2)/(1-q)+a1(1-1/4)/(1-q)
=9a1/[4*(1-q)]
2s9=2a1(1-q^9)/(1-q)
=2a1(1+1/8)/(1-q)
=9a1/[4*(1-q)]
2s9=s3+s6成等差数列