设等差数列的通项an=a1+(n-1)d,前n项的和sn=na1+n(n?1)d2,因为a5=3a7得到a1+4d=3(a1+6d),解得a1=-7d,代入到sn中得:sn=-d2n2-15d2n,当n=7或8时,Sn取得最大值.故答案为7或8.