{an}{bn}为等差数列,前n项和分别为Sn,Tn,(1)S(2n+1)/Tn=2n+1/n+4 求a10/b5(2)

∵a10=(a1+a19)/2
∴S(2*9+1)=s19=(a1+a19)*19/2=a10*19
∵b5=(b1+b9)/2
∴T9=(b1+b9)*9/2=b5*9
∵S19/T9=19/(9+4)=19/13
∴a10*19/(b5*9)=19/13
∴a10/b5=9/13