设等差数列{an}的前n项和为Sn,等差数列{bn}的前n项和为Tn,若Tn\Sn=4n+27\7n+1,求bn\an
问题描述：
设等差数列{an}的前n项和为Sn,等差数列{bn}的前n项和为Tn,若Tn\Sn=4n+27\7n+1,求bn\an
答
设等差数列{an}的公差为a
等差数列{bn}的公差为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)