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,S9,S6 成等差数列.

s3+s6=2s9

when q=1,a2,a8,a5 成等差数列.

when q不等于1；

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

约分得：a1q^3+a1q^6=2a1q^9

两边各约去q^2;

a1q+a1q^4=2a1q^7

即a2+a5=2a8;

所以命题成立