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