在等差数列{an}中,若a1003+a1004+a1005+a1006=18,则该数列的前2008项的和为(  ) A.18072 B.3012 C.9036 D.12048

问题描述:

在等差数列{an}中,若a1003+a1004+a1005+a1006=18,则该数列的前2008项的和为(  )
A. 18072
B. 3012
C. 9036
D. 12048

∵a1003+a1004+a1005+a1006
=(a1003+a1006)+(a1004+a1005
=2(a1+a2008)=18,
∴a1+a2008=9,
则S2008=

2008(a1+a2008
2
=1004(a1+a2008)=1004×9=9036.
故选C