已知等差数列{An}的前n项和Sn=2*n^2-15*n+a.求a的值；数列{An}的通项公式；使Sn取最小值的n

a(1)=s(1)=2-15+a=a-13,
a(2)=s(2)-s(1)=2*4-30+a-(a-13)=-9.
a(n+1)=s(n+1)-s(n)=2(n+1)^2-15(n+1)+a-2n^2+15n-a=4n+2-15=4n-13,
a(n)=4(n-1)-13,
-13=a(1)=a-13, a=0.
s(n)=2n(n-1)-13n=2n^2-15n=2[n^2-15n/2 + 225/16 - 225/16]=[n-15/4]^2 - 225/8,
s(3)=3(2*3-15)=-27,
s(4)=4(2*4-15)=-28.
s(n)>=s(4)=-28.
n=4时,s(n)取最小值.