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

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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=__

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'=t
a7/b7=(a1+6d)/(b1+6d')
=(t/2 +12t)/(2t+6t)
=25/16
an/bn=[a1+(n-1)d]/[b1+(n-1)d']
=[t/2 +2(n-1)t]/[2t+(n-1)t]
=(4n-3)/(2n+2)

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