题目内容
等差数列{an}共有2n+1项,其中奇数项之和为4,偶数项之和为3,则n的值是( )
A.3 | B.5 | C.7 | D.9 |
设数列公差为d,首项为a1,
奇数项共n+1项,其和为S奇=
=
=(n+1)an+1=4,①
偶数项共n项,其和为S偶=
=
=nan+1=3,②
得,
=
,解得n=3
故选A
奇数项共n+1项,其和为S奇=
(n+1)(a1+a2n+1) |
2 |
(n+1)2an+1 |
2 |
偶数项共n项,其和为S偶=
n(a2+a2n) |
2 |
n2an+1 |
2 |
① |
② |
n+1 |
n |
4 |
3 |
故选A
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