From S20=S 10 (1):

2a 1+29d=0，

∫a 1 = 29，

∴d=-2

∴an=29+(-2)(n- 1)=3 1-2n，

∴Sn=n(a 1+an) /2

=-n^2+30n

=-(n- 15)2+225，

∴ When n= 15, the maximum value of Sn is 225.

(2): from s20 = s10: a1+a12+…+A20 = 0,

That is 5(a 15+a 16)=0, ①.

∵a 1=29>0,∴a 15>0,a 16