已知等差数列{an}中,a1大于0,且sm=sk(m不等于k),问n为何值时sn最大

设公差为d.Sm=Sk
ma1+m(m-1)d/2=ka1+k(k-1)d/2
(m-k)a1+[m(m-1)-k(k-1)]d/2=0
(m-k)a1+[m²-k²-m+k]d/2=0
(m-k)a1+[(m+k)(m-k)-(m-k)]d/2=0
(m-k)a1+(m-k)(m+k-1)d/2=0
m≠k,等式两边同除以m-k
a1+(m+k-1)d/2=0
a1=-(m+k-1)d/2 a1>0 因此d