因为是等差数列an=[a1+a(2n-1)]/2bn=[b1+b(2n-1)]/2所以S(2n-1)=n*[a1+a(2n-1)]/2=n*anT(2n-1)=n*[b1+b(2n-1)]/2=n*bnan/bn=S(2n-1)/T(2n-1)=(4n-3)/(6n-3)选D
