> 数学 >
等差数列{an},{bn}的前n项和分别为Sn,Tn,若
Sn
Tn
2n+2
n+3
a7
b7
的值为______.
人气:131 ℃ 时间:2019-10-17 02:32:48
解答
由等差数列的性质可得
a7
b7
=
2a7
2b7
=
a1 +a13
b1+b13
=
13×
(a1+a13)
2
13(b1+b13)
2
=
S13
T13

Sn
Tn
2n+2
n+3

S13
T13
=
2×13+2
13+3
=
7
4

故答案为
7
4
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