两个等差数列{an}和{bn}，它们的前n项和分别为Sn和Tn，若SnTn=2n+53n-2，则a5+a6+a7b5+b6+b7=．

Question

两个等差数列{a_n}和{b_n}，它们的前n项和分别为S_n和T_n，若 (Sn)/(Tn)= (2n+5)/(3n-2)，则 (a5+a6+a7)/(b5+b6+b7)=___．

Answer 1

由题意和等差数列的求和公式和性质可得：
(a5+a6+a7)/(b5+b6+b7)=(3a6)/(3b6)=(2a6)/(2b6)=(a1+a11)/(b1+b11)=((11(a1+a11))/(2))/((11(b1+b11))/(2))
=(S11)/(T11)=(2×11+5)/(3×11-2)=(27)/(31)
故答案为：(27)/(31)