Question

等比数列{an}的各项均为正数,2an,a3,成等差数列,且a3=2a2的平方,求an通项公式

Answer 1

设等比数列的公比是q,则有a3=a1q^2,a4=a1q^3,a5=a1q^4

2a4,a3,4a5成等比数列,则有a3^2=2a4*4a5

(a1q^2)^2=8a1q^3*a1q^4=8a1^2q^7

q^3=1/8

q=1/2.

a3=2a2^2

a1q^2=2(a1q)^2=2a1^2q^2

得到a1=1/2.

故有an=a1q^(n-1)=(1/2)^n.