設等差數(shù)列{an}的公差為a
等差數(shù)列{bn}的公差為b
S(2n-1)=［a1+a(2n-1)］*(2n-1)/2
=2an*(2n-1)/2
=(2n-1)an
T(2n-1)=［b1+b(2n-1］*(2n-1)/2
=2bn*(2n-1)/2
=(2n-1)bn
S(2n-1)/T(2n-1)=［(2n-1)an］/［(2n-1)bn］
=an/bn
∴an/bn=[4(2n-1)+27]/{7(2n-1)+1]=(8n+23)/(14n-6)