> 数学 >
等差数列{an},{bn}的前n项和分别为Sn,Tn,且
Sn
Tn
=
3n−1
2n+3
,则
a8
b8
=______.
人气:307 ℃ 时间:2019-10-26 06:40:15
解答
2a8
2b8
a1+a15
b1+b15

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

=
S15
T15
=
3×15−1
2×15+3
=
4
3

故答案为:
4
3
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