∵{an}与{bn}是等差数列∴Sn=[n(a1+an)]/2Tn=[n(b1+bn)]/2∴Sn/Tn=(a1+an)/(b1+bn)∵等差数列{an}与{bn}的前n项和的比为2n:(3n+1)∴(a1+an)/(b1+bn)=2n:(3n+1)假设(n+1)/2 =k {(n+1)/2为项数}则n=2k-1则ak/bk = 2(2k...