Question

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

Answer 1

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)