等差数列an,bn,其前n项和分别记为Sn,Tn,且Sn/Tn=(2n+5)/(2n-3)求a6/b6和an/bn

Question

等差数列an,bn,其前n项和分别记为Sn,Tn,且Sn/Tn=(2n+5)/(2n-3)
求a6/b6和an/bn

Answer 1

n=11
S11/T11=27/19
所以[(a1+a11)*11/2]/[(b1+b11)*11/2]=27/19
(a1+a11)/(b1+b11)=27/19
2a6/2b6=27/19
a6/b6=27/19
同样
S(2n-1)/T(2n-1)=(4n+3)/(4n-5)
则{[a1+a(2n-1)]*(2n-1)/2}/{[b1+b(2n-1)]*(2n-1)/2}=(4n+3)/(4n-5)
[a1+a(2n-1)][b1+b(2n-1)]=(4n+3)/(4n-5)
2an/2bn=(4n+3)/(4n-5)
an/bn=(4n+3)/(4n-5)