Question

已知等比数列(an)满足2a1+a3=3a2且a3+2是a2，a4的等差中项 求数列(an)的通项公式？

Answer 1

an=2^n
步骤：
等比数列{an}, =>an=a1*q^(n-1),(a1、q不为0)
=> a2=a1q,a3=a1q^2,a4=a1q^3,
2a1+a3=3a2
=>2a1+a1q^2=3a1q, => q^2-3q+2=0, =>q=1或2,
a3+2是a2，a4的等差中项, =>2(a3+2)=a2+a4, =>2a1q^2+4=a1q+a1q^3,
1) q=1时,=> 2a1+4=a1+a1,=>不成立。
2) q=2时, => 2a1*4+4=a1*2+a1*8, => a1=2,
=> an=a1*q^(n-1)=2*2^(n-1)=2^n.

Answer 2

2a1+a1*q²=3a1*q，q²-3q+2=0 得q=1或q=2，取q=2；a3+2=a2，a1*2²+2=a1*2，a1=-1
数列(an)的通项公式an=-2^(n-1)