题目内容

已知等差数列{an}的前n项和为Sn,若
OB
=a1
OA
+a2009
OC
,且A,B,C三点共线(O为该直线外一点),则S2009=______.
A,B,C三点共线得
AB
=t
AC
,所以
OB
=
OA
+
AB
=
OA
+t
AC
=
OA
+t(
OC
-
OA
)=(1-t)
OA
+t
OC

OB
=a1
OA
+a2009
OC
得1-t=a1,t=a2009,所以a1+a2009=1;
而sn=
n(a1+an
2
得s2009=
2009
2

故答案为
2009
2
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