> 数学 >
若两个等差数列{an},{bn}的前n项和分别为Sn,Tn,且满足
Sn
Tn
=
3n+2
4n−5
,则
a7
b7
=______.
人气:467 ℃ 时间:2019-10-19 13:37:52
解答
由题意可得
S14
T14
=
14(a1+a14)
2
14(b1+b14)
2
=
2a7
2b7
=
a7
b7
=
3×14+2
4×14−5
=
44
51

故答案为:
44
51
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