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成等比数列

Answer 1

首项a

2S9=2a(q^9-1)/(q-1)

S3+S6=a(a^3-1)/(q-1)+a(a^6-1)/(q-1)

2S9=S3+S6

显然a不等于0

2(q^9-1)=a^3-1+q^6-1

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

2q^7=q+q^4

2aq^7=aq+aq^4

2a8=a2+a5

所以a2,a8,a5成等差数列