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
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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!)
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!)
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
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
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