设等差数列{an}的公差为d,则 an=a1+(n-1)d,Sn=na1+[n(n-1)/2]d ∵a5=-13,S4=-82 ∴a1+4d=-13,4a1+6d=-82 解得a1=-25,d=3 ∴Sn=-25n+3[n(n-1)/2] 即Sn=(3/2)n^2-(53/2)n ∴S6=-105 ∵对于二次函数Sn=(3/2)n^2-(53/2)n,其对称轴为n=53/6,且(3/2)>0 ∴当n=9时,Sn取得最小值,且最小值为-117