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

2s9=s3+s6
2a1(1-q^9)/(1-q)=a1(1-q^3)/(1-q)+a1(1-q^6)/(1-q)
2(1-q^9)=(1-q^3)+(1-q^6)
2-2q^9=1-q^3+1-q^6
2q^9=q^3+q^6(同时除以q^2)
q^7=q+q^4(同时乘以a1)
2a1q^7=a1q+a1q^4
2a8=a2+a5
所以a2,a8,a5成等差数列