等差数列{an}和{bn}的前几项和分别为Sn和Tn,若Sn:Tn=2n:(3n+1),求a7:b7
问题描述:
等差数列{an}和{bn}的前几项和分别为Sn和Tn,若Sn:Tn=2n:(3n+1),求a7:b7
答
其实很容易.S13:T13=26:40=13:20
又a1+a13=2a7.S13=a1+---+a13=13a7
同理,T13=13b7
所以,S13:T13=26:40=13:20=13a7:13b7=a7:b7