> 数学 >
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
=______.
人气:210 ℃ 时间:2019-11-22 19:56:09
解答
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
推荐
猜你喜欢
© 2024 79432.Com All Rights Reserved.
电脑版|手机版