问题 填空题
两个等差数列an的和bn的前n项和分别为Sn和Tn,已知
Sn
Tn
=
5n-9
n+3
,则使an=tbn成立的正整数t的个数是 ______;
答案

5n-9
n+3
=1即n=3时,
S3
T3
=
a1+a2+a3
b1+b2+b3
=
3a2
3b2
=
a2
b2
=1,则a2=b2,此时t=1;

5n-9
n+3
=2即n=5时,
S5
T5
=
a1+a2+…+a5
b1+b2+…+b5
=
5a3
5b3
=
a3
b3
=2,则a3=2b3,此时t=2;

5n-9
n+3
=3即n=9时,
S9
T9
=
a1+a2+…+a9
b1+b2+…+b9
=
9a5
9b5
=
a5
b5
=3,则a5=3b5,此时t=3;

5n-9
n+3
=4即n=21时,
S21
T21
=
a1+a2+…+a21
b1+b2+…+b21
=
21a11
21b11
=
a11
b11
=4,则a11=4b11,此时t=4.

5n-9
n+3
≥5时,解得的n不为正整数即t也不为正整数,所以满足题意的正整数t的个数是4

故答案为:4

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