Question

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.

Answer 1

S3+S6=2S9 S3+q^3S3+S3=2(1+q^3+q^6)S3

2+q^3=2+2q^3+2q^6 2q^6=-q^3 q^3=-1/2

a2(1+q^3)=1/2a2 2a8=2a2*q^6=1/2a2 a2+a5=2a8 即得证