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

an = a1q^(n-1)
a1,(1/2)a3,2a2成等差数列
2a2+a1= a3
2a1q+a1=a1q^2
q^2-2q-1=0
q=1+√2
(a9+a10)/(a7+a8)
=a1q^8(1+q)/[a1a^7(1+q)]
=q
=1+√2