已知各项均为正数的等比数列{an}中,3a1,1/2a3,2a2成等差数列,则(a11+a13)/(a8+a10)=
问题描述:
已知各项均为正数的等比数列{an}中,3a1,1/2a3,2a2成等差数列,则(a11+a13)/(a8+a10)=
答
a1q^2=3a1+2a1q、q^2-2q-3=0、q=-1(舍去)、q=3.
(a11+a13)/(a8+a10)=(a1q^10+a1q^12)/(a1q^7+a1q^9)=q^3=27.(a1q^10+a1q^12)/(a1q^7+a1q^9)=q^3=27。怎么得到的q^3(a1q^10+a1q^12)/(a1q^7+a1q^9)=q^10(a1+a1q^2)/[q^7(a1+a1q^2)=q^10/q^7=q^3