Question

在等比数列{an}中,a2-a1=2,且2a2为3a1和a3的等差中项,求数列{an}的首项,公比及前n项和

Answer 1

3a1,2a2,a3成等差数列有4a2=3a1+a3
又 (a2)²=a1×a3
所以有(a2)²=a1×（4a2-3a1)=4a1a2-3a1^2
(a2-3a1)(a2-a1)=0
所以a2=3a1 或 a2=a1

又有a2-a1=2,故a2=a1不成立,舍,即有a2=3a1
3a1-a1=2
a1=1
公比q=a2/a1=3
an=a1q^(n-1)=3^(n-1)
Sn=1*(3^n-1)/(3-1)=1/2*(3^n-1)

Answer 2

an= a1q^(n-1)
a2-a1=2
a1(q-1)=2 (1)
2a2为3a1和a3的等差中项
4a2=3a1+a3
4(a1q)=a1(3+q^2)
q^2-4q+3=0
(q-1)(q-3)=0
q=1 (rejected) or 3
a1=1
an = a1q^(n-1) = 3^(n-1)
Sn = (3^n -1)/2