Question

设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

Answer 1

S3,S9,S6成等差:2*S9=S3+S6

2*a1(1-q^9)/(1-q)=a1(1-q^3)/(1-q)+a1(1-q^6)/(1-q)

2*q^9=q^3+q^6

2*q^6=1+q^3

同乘a1*q：2*a1*q^7=a1*q+a1*q^4

即2*a8=a2+a5,即证