已知等比数列{an}中,各项都是正数,且a1,1/2a3,2a2成等差数列,则a9+a10/a7+a8的值为?

Question

已知等比数列{an}中,各项都是正数,且a1,1/2a3,2a2成等差数列,则a9+a10/a7+a8的值为?

Answer 1

a1,1/2a3,2a2
a1+2a2=a3
a1+2a1q=a1q^2
1+2q=q^2
q^2-2q-1=0
(q-1)^2=2
q1=1+√2
q2=1-√2
等比数列{an}中,各项都是正数,所以q=1+√2
a9+a10/a7+a8
=(q^8+q^9)/(q^6+q^7)
=(q^2+q^3)/(1+q)
=q^2
=3+2√2