等差数列{an},{bn}的前n项和分别为Sn和Tn,若Sn/Tn=(7n+2)/(n+3),求a6/b6
问题描述：
等差数列{an},{bn}的前n项和分别为Sn和Tn,若Sn/Tn=(7n+2)/(n+3),求a6/b6
答
Sn/Tn=(7n+2)/(n+3) S(2n-1)/T(2n-1)=[7(2n-1)+2]/(2n-{[a1+a(2n-1)]*(2n-1)/2}/2an/(2bn)=(14n-5)/(2n+2) a
答
Sn/Tn=(7n+2)/(n+3)
S(2n-1)/T(2n-1)=[7(2n-1)+2]/(2n-1+3)
S(2n-1)/T(2n-1)=(14n-5)/(2n+2)
{[a1+a(2n-1)]*(2n-1)/2}/{[b1+b(2n-1)]*(2n-1)/2}=(14n-5)/(2n+2)
[a1+a(2n-1)]/[b1+b(2n-1)]=(14n-5)/(2n+2)
2an/(2bn)=(14n-5)/(2n+2)
an/bn=(14n-5)/(2n+2)
a6/b6=(14*6-5)/(2*6+2)
a6/b6=79/14