Question

若等比数列｛an｝的前n项和为Sn，a2=6，S3=21，则公比q为？

Answer 1

题目应该为S3，S9，S6成等差数列证明：首先根据等差数列的通项公式an=a1+（n-1）d则a(n+1)=a1+（n)d
a（n+2)=a1+（n+1）d很显然等差数列有an+a(n+2)=2a(n+1)根据等比数列的求和公式：Sn=a1/(1-q)-a1/(1-q)*q^n
则S3=a1/(1-q)-a1/(1-q)*q^3
S9=a1/(1-q)-a1/(1-q)*q^9S6=a1/(1-q)-a1/(1-q)*q^6因为S3，S9，S6成等差数列即有S3+S6=2S9即a1/(1-q)-a1/(1-q)*q^3+a1/(1-q)-a1/(1-q)*q^9=2[a1/(1-q)-a1/(1-q)*q^6]可以化为
a1q^1+a1q^7=a1q^4--------(1)根据等比数列的通项公式a2=a1q^1
a8=a1q^7a5=a1q^4
分别代入(1)式中得到：a2+a8=2a5再根据前面已经给出的等差数列有an+a(n+2)=2a(n+1)则a2,a8，a5成等差数列
得证。