设数列{an}是公差部位0的等差数列,前n项和为110,且a1,a2,a4,成等比数列,求an的通项公式

等差数列前n项和为：Sn=na1+n(n-1)d/2=110.a1,a2,a4成等比数列,则a2^2=a1a4,即(a1+d)^2=a1(a1+3d),解此式得：a1=d.代入Sn式中得：na1+n(n-1)a1/2=110,得：a1=220/[n(n+1)]所以：an=a1+(n-1)d=a1+(n-1)a1=na1=220/(n+...