已知,a1+a3+a5+...+a99=80, d=3
由:an=a1+(n-1)d
可得:a1+(a1+2d)+(a1+4d)+...+(a1+98d)=80
∴50a1+2450d=80
∵d=3
∴50a1+7350=80
a1=-145.4
∴前一百项就是:a1+a2+a3+...+a100
=100a1+(d+2d+3d+...+99d)
=310
a1+a3+a5+…+a99=a1+a1+2d+a1+4d+…+a1+98d=50a1+24*100d+50d=50a1+2450d=80
50a1+2450*3=80
50a1=-7270
a1=-145.4
S100=100a1+[100*(100-1)*d]/2
=100*(-145.4)+(100*99*3)/2
=-14540+14850
=310