an/bn=S(2n-1)/T(2n-1)证明过程：设等差数列{an}，前n项和Sn，等差数列{bn}，前n项和Tn。an/bn ={[a1+a(2n-1)]/2}/{[b1+b(2n-1)]/2} （等差中项性质）={[a1+a(2n-1)](2n-1)/2}/{[b1+b(2n-1)](2n-1)/2} （分子分母同乘以2n-1）=S(2n-1)/T(2n-1) （...
=>an/bn=(4n+1)/(6n+2)=>a1/b1=5/8
=>an/bn=(4n+1)/(6n+2)=>a1/b1=5/8