若两个等差数列{an}和{bn}的前n项和分别为Sn和Tn,且满足SnTn=7n+1/n+3,则a2+a5+a17+a22b8+b10+b12+b16=_.

问题描述:

若两个等差数列{an}和{bn}的前n项和分别为Sn和Tn,且满足

Sn
Tn
7n+1
n+3
,则
a2+a5+a17+a22
b8+b10+b12+b16
=______.

由等差数列的通项公式可得

a2+a5+a17+a22
b8+b10+b12+b16
=
2(2a1+21d)
2(2b1+21d′)
=
a1+a22
b1+b22
=
22(a1+a22)
2
22(b1+b22)
2
=
S22
T22
=
7×22+1
22+3
=
155
25
=
31
5

故答案为
31
5