Question

等比数列{An}的前n项和为Sn，已知S1.S3.S2成等差数列 1，求{An}的公比q 2.求a1－a2=3,求Sn

Answer 1

（1）S1.S3.S2成等差数列
则公差d=S3-S1=a2+a3
=S2-S3=-a2
于是a2+a3=-a2,a3/a2=-2
{an}为等比数列，则q=-2
（2）a1－a2=a1-a1*q=3a1=3
a1=1
Sn=a1*(1-q^n)/(1-q)
=[1-(-2)^n]/[1-(-2)]
=[1-(-2)^n]/3

Answer 2

S1=a1
S2=a1(1+q)
S3=a1(1+q+q^2)
S1,S3,S2成等差数列
即
s3-s1=s2-s3
1+q+q^2-1=1+q-(1+q+q^2)
q^2+q=-q^2
q=0（舍去）或-1/2

a1(1-1/4)=3
a1=4
所以Sn=4*(1-(-1/2)^n)/(3/2)=8*(1-(-1/2)^n)/3

Answer 3

(1)S3-S1=S2-S3
a1+a2+a3-a1=a1+a2-a1-a2-a3
a2+2a3=0
a2+2a2q=0
q=-1/2
(2)a1-a3=3
a1-a1/4=3
a1=4
Sn=a1(1-q^n)/(1-q)=4[1-(-1/2)^n]/(3/2)=8[1-(-1/2)^n]/3

Answer 4

1）S1.S3.S2成等差数列
则公差d=S3-S1=a2+a3
=S2-S3=-a3
于是a2+a3=-a2,a3/a2=-1/2
{an}为等比数列，则q=-1/2
（2）a1－a2=a1-a1*q=3a1=3
a1=1
Sn=a1*(1-q^n)/(1-q)
=[1-(-2)^n]/[1-(-2)]
=[1-(-2)^n]/3