若等差数列{an}{bn}的前n项和为Sn和Tn,且Sn/Tn=(2n-1)/(n+3),则a7/b7=___an/bn=__
问题描述：
若等差数列{an}{bn}的前n项和为Sn和Tn,且Sn/Tn=(2n-1)/(n+3),则a7/b7=___an/bn=__
答
设数列{an}公差为d,数列{bn}公差为d'.Sn/Tn=[na1+n(n-1)d/2]/[nb1+n(n-1)d'/2]=[dn+(2a1-d)]/[d'n+(2b1-d')]=(2n-1)/(n+3)令d=2t,则2a1-d=-t,d'=t,2b1-d'=3t解得a1=t/2 d=2t b1=2t d'=ta7/b7=(a1+6d)/(b1+6d')=(t/2 ...