a1=-11
a4=a1+3d,a6=a1+5d
a4a6=(a1+3d)(a1+5d)=(a1)^2+8a1d+15d^2=-6
即15d^2-88d+127=0因为a4a6<0,a1<0,于是a4<0,a6>0,于是sn最小值在s4与s5之中于是-11+3d<0,-11+5d>011/5<d<11/315d^2-88d+127=0对称轴=44/15a5=a1+4d,a5<0,即d<11/4令d=11/415(11/4)^2-88(11/4)+121+6>15(11/4)^2-88(11/4)+121>121[(15/16)-2+11]>0【结合函数思想,于零点原理】于是d<11/4于是a5<0那么sn最小值为s5
设等差数列an的前n项和为Sn,若a1=-11,a4a6=-6,则当Sn取最小值时,n
设等差数列an的前n项和为Sn,若a1=-11,a4a6=-6,则当Sn取最小值时,n
等于
等于
数学人气:441 ℃时间:2020-04-15 12:29:00
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