已知两个等差数列{An} {bn},他们的前n项和分别是Sn,Tn ,若Sn/Tn=3n-1/2n+3,求a9/b7

设{an}公差为d,{bn}公差为d'
Sn/Tn=[na1+n(n-1)d/2]/[nb1+n(n-1)d'/2]
=[2a1+(n-1)d]/[2b1+(n-1)d']
=[dn+(2a1-d)]/[d'n+(2b1-d')]
=(3n-1)/(2n+3)
令d=3t,则2a1-d=-t,d'=2t,2b1-d'=3t
解得a1=t d=3t b1=(5/2)t d'=2t
a9/b7=(a1+8d)/(b1+6d')
=(t+8×3t)/[(5/2)t+6×(2t)]
=50/29
知识拓展：
推广：
am/bn=[a1+(m-1)d]/[b1+(n-1)d']
=[t+(m-1)(3t)]/[(5/2)t+(n-1)(2t)]
=(6m-4)/(4n +1)