已知Sn是等比数列an的前n项和,S3,S9,S6成等差数列.求公比q.并证明ak,a（k+6）,a（k+3）成等差数列

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/05/18 04:06:49

{an}为等比数列
S3,S9,S6成等差数列
那么2S9=S3+S6
若q=1,则Sn=na1 ,a1≠0
则18a1=3a1+6a1,不合题意
则q≠1
∴2a1(q^9-1)/(q-1)=a1(q^3-1)/(q-1)+a1(q^6-1)/(q-1)
==>
2（q^9-1)=q^3-1+q^6-1
∴2q^9-q^6-q^3=0
2q^6-q^3-1=0
(2q^3+1)(q^3-1)=0
解得q^3=1(舍去）或q^3=-1/2
∴q=-1/³√2
2a(k+6)=2a1*q^(k+5)
=2a1*q^(k-1)*q^6
=2*(-1/2)^2*a1*q^(k-1)
=1/2*a1*q^(k-1)
ak+a(k+3)
=a1*q^(k-1)+a1*q^(k+2)
=a1*q^(k-1)+a1*q^(k-1)*q^3
=a1*q^(k-1)-1/2*a1*q^(k-1)
=1/2a1*q^(k-1)
2a（k+6）=ak+a（k+3）
即ak,a（k+6）,a（k+3）为等差数列.

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