(I)设等差数列{an}的公差为d≠0,利用成等比数列的定义可得,,再利用等差数列的通项公式可得,化为d(2a1+25d)=0,解出d即可得到通项公式an;
(II)由(I)可得a3n-2=-2(3n-2)+27=-6n+31,可知此数列是以25为首项,-6为公差的等差数列.利用等差数列的前n项和公式即可得出a1+a4+a7+…+a3n-2.
【解析】
(I)设等差数列{an}的公差为d≠0,
由题意a1,a11,a13成等比数列,∴,
∴,化为d(2a1+25d)=0,
∵d≠0,∴2×25+25d=0,解得d=-2.
∴an=25+(n-1)×(-2)=-2n+27.
(II)由(I)可得a3n-2=-2(3n-2)+27=-6n+31,可知此数列是以25为首项,-6为公差的等差数列.
∴Sn=a1+a4+a7+…+a3n-2=
=
=-3n2+28n.