a(n)=a+(n-1)d,
d>0时,a(n)单调增,
ds(9)=9[a+4d],0a+4d,0< a+9d/2,
d/2 > a+4d+d/2 = a+9d/2 > 0.
d>0.a(n)单调增.
-9d/2 < a < -4d < 0.
a(5)=a+4da+9d/2>0.
a(1)0,s(6)>s(5).
因此,总有,
n不为5时,s(n)>s(5).
n=5时,s(n)取最小值.