等差数列{an}、{bn}的前n项和分别为Sn、Tn,且SnTn=7n+45n−3,则使得anbn为整数的正整数的n的个数是(  ) A.3 B.4 C.5 D.6

问题描述:

等差数列{an}、{bn}的前n项和分别为Sn、Tn,且

Sn
Tn
7n+45
n−3
,则使得
an
bn
为整数的正整数的n的个数是(  )
A. 3
B. 4
C. 5
D. 6

∵等差数列{an}、{bn},
∴an=

a1+a2n−1
2
,bn=
b1+b2n−1
2

an
bn
=
nan
nbn
=
n(a1+a2n−1)
2
n(b1+b2n−1)
2
=
S2n−1
T2n−1
,又
Sn
Tn
=
7n+45
n−3

an
bn
=
7(2n−1)+45
(2n−1)−3
=7+
66
2n−4

经验证,当n=1,3,5,13,35时,
an
bn
为整数,
则使得
an
bn
为整数的正整数的n的个数是5.
故选C