两个等差数列{an}和{bn},其前n项和分别为Sn,Tn,且SnTn=7n+2n+3,则a2+a20b7+b15等于(  ) A.94 B.378 C.7914 D.14924

问题描述:

两个等差数列{an}和{bn},其前n项和分别为Sn,Tn,且

Sn
Tn
7n+2
n+3
,则
a2+a20
b7+b15
等于(  )
A.
9
4

B.
37
8

C.
79
14

D.
149
24

因为:

a2+a20
b7+b15
=
a1+a21
b1+b21

=
21
2
(a1+a21)
21
2
(b1+b21

=
S21
T21
=
7×21+2
21+3
=
149
24

故选:D.