问题 填空题
已知两等差数列{an},{bn}的前n项和分别为Sn,Tn,且
Sn
Tn
=
2n+1
n+2
,则
a8
b7
=______.
答案

设Sn=kn(2n+1),Tn=kn(n+2),(k≠0),

∵数列{an},{bn}是等差数列,

∴an=3k+4k(n-1)=4kn-k,bn=3k+2k(n-1)=2kn+k,

a8
b7
=
32k-k
14k+k
=
31
15

故答案为

31
15

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