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

由题意易知 a3 = a1 + 2a2a1 * q^2 = a1 + 2a1 * q （a1不等于0）即 q^2 - 2q - 1 = 0 ,解得 q = 1 + √2 或 -1 + √2 （√2 指根号2）(a9+a10)/(a7+a8) = a9(1+q)/[a7(1+q)] = a9/a7= q^2= 3 + 2√2 或 3 - 2√2...