等差数列{an},{bn}的前n项和分别为Sn,Tn,若Sn/Tn＝2n/3n+1,则a5/a7=
问题描述：
等差数列{an},{bn}的前n项和分别为Sn,Tn,若Sn/Tn＝2n/3n+1,则a5/a7=
答
应该是求a5/b7吧
根据等差数列前n项和的性质：设Sn=2kn²,Tn=kn(3n+1)
则a5/b7
=(S5-S4)/(T7-T6)
=(2k×5²-2k×4²)/(k×7×22-k×6×19)
=(18k)/(40k)
=9/20
答案：9/20