Question

等比数列｛An}前n项和Sn已知S1S3S2成等差求｛An｝的公比q

第二问：求a1-a3=3求Sn

Answer 1

1、S1=a1
S2 =a1+a2= a1(1+q)
S3 =a1+a2+a3= a1(1+q+q^2)
因S1+S2 = 2S3
则2a1+a1q = 2a1(1+q+q^2)
因为是等比数列a1≠0 q≠0
所以2+q=2+2q+2q^2
2q^2+q=0
解得q =-1/2
2、因a1- a3=a1-a1*q^2 =a1*(1-1/4)=3
所以a1=4
an=a1*q^(n-1)=4*(-1/2)^(n-1)
Sn =4*[1-(-1/2)^n]/(1+1/2)
= (8/3)[1-1/2^n]
希望能帮到你，祝学习进步O(∩_∩)O

Answer 2

an=a1q^((n-1)，2S3=S1+S2
q=1时：a1=0,不合题意，因为等比数列定义时a1,q不为零
q不等于1，
2S3=2*a1(1-q^3)/(1-q)=S1+S2=a1+a1(1-q^2)/(1-q)，a1，q不等于零
2q^3=q^2+q
2q^2=q+1
q1,2=1（舍掉），-1/2

（2）q=-1/2，
a1-a3=a1(1-q^2)=3a1/4=3
故a1=4
于是Sn=a1(1-q^n)/(1-q)=8/3(1-(-1/2)^2)

Answer 3

1、
S1=a1
S2 = a1(1+q)
S3 = a1(1+q+q^2)
S1+S2 = 2S3
a1(2+q) = 2a1(1+q+q^2)
解得q =-2

2、
a1 - a1*q^2 = -3a1 = 3
a1=-1
Sn =