题目内容
在等差数列{an}中,a2+3a7=0,且a1>0,Sn是它的前n项和,当Sn取得最大值时的n=______.
∵等差数列{an}中,a2+3a7=0,a1>0,
∴(a1+d)+3(a1+6d)=0,
a1=-
d,d<0,
∴Sn=na1+
d
=-
n+
n2-
n
=
(n2 -
n)
=
(n-
)2-
,
∴n=5时,Sn取得最大值.
故答案为:5.
∴(a1+d)+3(a1+6d)=0,
a1=-
| 19 |
| 4 |
∴Sn=na1+
| n(n-1) |
| 2 |
=-
| 19d |
| 4 |
| d |
| 2 |
| d |
| 2 |
=
| d |
| 2 |
| 21 |
| 2 |
=
| d |
| 2 |
| 21 |
| 4 |
| 441d |
| 32 |
∴n=5时,Sn取得最大值.
故答案为:5.
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