题目内容
在等差数列{an}中,a1=-2 012,其前n项和为Sn,若
-
=2,则S2012的值等于( )
| S12 |
| 12 |
| S10 |
| 10 |
| A.-2 011 | B.-2 012 | C.-2 010 | D.-2 013 |
∵数列{an}为等差数列,设其公差为d,则其前n项和为Sn=na1+
d,
∴
=a1+
d,
∴
-
=
,
∴{
}为公差是
的等差数列,
∴
-
=2×
=d,
又
-
=2,
∴d=2.
∵数列{an}为等差数列,a1=-2 012,
∴S2012=2012a1+
d
=2012×(-2012)+
×2
=-2012.
故选B.
| n(n-1) |
| 2 |
∴
| Sn |
| n |
| (n-1) |
| 2 |
∴
| Sn+1 |
| n+1 |
| Sn |
| n |
| d |
| 2 |
∴{
| Sn |
| n |
| d |
| 2 |
∴
| S12 |
| 12 |
| S10 |
| 10 |
| d |
| 2 |
又
| S12 |
| 12 |
| S10 |
| 10 |
∴d=2.
∵数列{an}为等差数列,a1=-2 012,
∴S2012=2012a1+
| 2012×(2012-1) |
| 2 |
=2012×(-2012)+
| 2012×(2012-1) |
| 2 |
=-2012.
故选B.
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