两个等差数列{an}和{bn}的前n项和分别为Sn和Tn,若SnTn=7n+3/n+3,则a8b8=_.

问题描述:

两个等差数列{an}和{bn}的前n项和分别为Sn和Tn,若

Sn
Tn
7n+3
n+3
,则
a8
b8
=______.

Sn
Tn
7n+3
n+3

a8
b8
2a8
2b8
=
a1+a15
b1+b15
=
15
2
(a1+a15)
15
2
(b1+b15)

=
S15
T15
=
7×15+3
15+3
=6
故答案为:6