两等差数列{an},{bn}的前n项和分别为Sn,Tn,若SnTn=2n+33n+1,则a7b7=()A.3346B.1722C.2940D.3143-高二数学

题目简介

两等差数列{an},{bn}的前n项和分别为Sn,Tn,若SnTn=2n+33n+1,则a7b7=()A.3346B.1722C.2940D.3143-高二数学

题目详情

两等差数列{an},{bn}的前n项和分别为Sn,Tn,若
Sn
Tn
=
2n+3
3n+1
,则
a7
b7
=(  )
A.
33
46
B.
17
22
C.
29
40
D.
31
43
题型:单选题难度:偏易来源:不详

答案

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
=class="stub"2×13+3
3×13+1
=class="stub"29
40

故选C

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