设数列{an}首项为a1,公差为d；数列{bn}首项为b1,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)]/[nd'+(2b1-d)]
=(2n+1)/(n+2)
令d=2t,则2a1-d=t,d'=t,2b1-d=2t
解得
a1=3t/2 d=2t b1=3t/2 d'=t
an/bn=[a1+(n-1)d]/[b1+(n-1)d']=[(3t/2)+2(n-1)t]/[(3t/2)+(n-1)t]=(4n-1)/(2n+1)
a5/b5=(4×5-1)/(2×5+1)=19/11
提示：用这种方法可以求数列{an}中任意项与数列{bn}中任意项的比值.