> 数学 >
等差数列{an}与{bn}的前n项和分别为Sn与Tn,若
Sn
Tn
=
3n-2
2n+1
,则
a7
b7
=(  )
A.
37
27

B.
38
28

C.
39
29

D.
40
30
人气:143 ℃ 时间:2019-10-24 13:39:20
解答
由等差数列的性质可得:
a7
b7
=
13a7
13b7

=
13×
a1+a13
2
13×
b1+b13
2
=
S13
T13
=
3×13-2
2×13+1
=
37
27

故选A
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