(1)设等差数列{an}的公差为d,由题意得,S17=S9,即 a10+a11+…+a17= 8(a10+a17) 2 =0,∴2a1+25d=0,又a1=25,解得d=-2,∴an=27-2n,(2)由(1)得,Sn= n(a1+an) 2 = n(25+27?2n) 2 =-n2+26n=169-(n-13)2,∴当n=13时,Sn最大,且Sn的最大值为169.