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+1n+3，则a2+a5+a17+a22b8+b10+b12+b16=．

Question

若两个等差数列{a_n}和{b_n}的前n项和分别为S_n和T_n，且满足(Sn)/(Tn)＝(7n+1)/(n+3)，则(a2+a5+a17+a22)/(b8+b10+b12+b16)=______．

Answer 1

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)．