问题
选择题
设Sn是等差数列{an}的前n项和,若
|
答案
设等差数列{an}的首项为a1,由等差数列的性质可得
a1+a9=2a5,a1+a5=2a3,
∴
=s9 s5
=
×9a1+a9 2
×5a1+a5 2
=9a5 5a3
×9 5
=1,5 9
故选A.
设Sn是等差数列{an}的前n项和,若
|
设等差数列{an}的首项为a1,由等差数列的性质可得
a1+a9=2a5,a1+a5=2a3,
∴
=s9 s5
=
×9a1+a9 2
×5a1+a5 2
=9a5 5a3
×9 5
=1,5 9
故选A.