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

题目

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

答案

{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-...