Question

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

Answer 1

{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.