设等差数列{an}，{bn}的前n项和分别为Sn，Tn若对任意自然数n都有SnTn=2n-34n-3，则a9b5+b7+a3b8+b4的值为．

Question

设等差数列{a_n}，{b_n}的前n项和分别为S_n，T_n若对任意自然数n都有(Sn)/(Tn)=(2n-3)/(4n-3)，则(a9)/(b5+b7)+(a3)/(b8+b4)的值为 ___ ．

Answer 1

由等差数列的性质和求和公式可得：
(a9)/(b5+b7)+(a3)/(b8+b4)=(a9)/(b1+b11)+(a3)/(b1+b11)
=(a3+a9)/(b1+b11)=(a1+a11)/(b1+b11)=((11(a1+a11))/(2))/((11(b1+b11))/(2))
=(S11)/(T11)=(2×11-3)/(4×11-3)=(19)/(41)
故答案为：(19)/(41)