等差数列{an}中,Sn是其前n项和,a1=2008,S2007/2007-S2005/2005=-2,S2009=?

Sn=na1+n(n-1)d/2,则S2007/2007-S2005/2005=a1+1003d-a1-1002d=d=-2,
S2009=2009a1+2009(2009-1)d/2=2009×2008+2009×2008×(-2)/2=0

S2009=0

用等差中项，S2007＝2007a1004，S2005＝2005a1003

an=a1+(n-1)*d
sn=(a1+an)*n/2=a1*n+[n*（n-1）*d]/2
s2007=2008*2007+2007*2006d/2
a2005=2008*2005+2005*2004d/2
s2007/2007=(2008*2007+2007*2006d/2)/2007=2008+1003d
s2005/2005=(2008*2005+2005*2004d/2)/2005=2008+1002d
所以d=-2,
s2009=2008*2009+2009*2008*(-2)/2=0