若一等差数列前四项的和为124，后四项的和为156，又各项的和为350，则此数列共有（ ）
A. 10项
B. 11项
C. 12项
D. 13项
人气：477 ℃ 时间：2020-05-16 07:45:57
解答
设此等差数列为{a
n}，则a
1+a
n=a
2+a
n-1=a
3+a
n-2=a
4+a
n-3．
∵a
1+a
2+a
3+a
4=124，a
n-3+a
n-2+a
n-1+a
n=156，
∴4（a
1+a
n）=124+156，解得a
1+a
n=70．
又
Sn＝=
=350，
解各n=10．
故选A．
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