设等差数列{an}{bn}的前n项和分别为Sn,Tn,若对任意n∈N﹡都有Sn/Tn=2n-3/4n-3,则a9/(b5+b7)+_数学解答

设等差数列{an}{bn}的前n项和分别为Sn,Tn,若对任意n∈N﹡都有Sn/Tn=2n-3/4n-3,
则a9/(b5+b7)+a3/(b4+b8)为多少,

人气：155 ℃　时间：2020-01-30 11:39:04

解答

因为在等差数列中,所以b5+b7=b4+b8
所以a9/(b5+b7)+a3/(b4+b8)= (a9+a3)/(b4+b8)
=(a1+a11)/(b1+b11)
=（a1+a11)x11/2/(b1+b11)x11/2=S11/T11
=19/41