Question

关于等差数列基础题已知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（步骤

Answer 1

设等比数列首项A1,公比是Q(Q1)

S3=A1(1+Q+Q^2+Q^3)=A1(1-Q^3)/(1-Q)

S9=A1(1-Q^9)/(1-Q)

S6=A1(1-Q^6)/(1-Q)

S3,S9,S6成等差数列

S3+S6=2S9

A1(1-Q^3)/(1-Q)+A1(1-Q^6)/(1-Q)=2A1(1-Q^9)/(1-Q)

Q^3+Q^6=2Q^9 1+Q^3-2Q^6=0

2A8-A2-A5=A1(2Q^7-Q-Q^4)=A1Q(2Q^6-1-Q^3)=0

所以A2,A8,A5成等差数列