Question

设sn是等差数列an的前n项和,且a7=4

问一道数列题目.
设Sn是等差数列an的前N项和 且3a4=7a7
且a1大于0 当sn取到最大值时,求n是多少?

Answer 1

{an}为等差数列
所以an=a1+(n-1)d
3a4=7a7
3(a1+3d)=7(a1+6d)
4a1=-33d
d=-4a1/33
a1=-33d/4
a1>0所以d