（I）设等差数列{an}的公差为d≠0,
由题意a1,a11,a13成等比数列,∴
a 211
=a1a13,
∴(a1+10d)2=a1(a1+12d),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=n(a1+a3n-2) /2= n(25-6n+31) /2
=-3n2+28n．