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由題意可得S12-S9=a10+a11+a12=3a11=0,∴a11=0又∵等差數列{an}中a1<0,∴{an}為等差數列,且前10項為負數,第11項為0,從第11項開始為正值,∴數列的前10項或前11項和最小,即使Sn最小的序號n為10或11,故答案為:10或11.
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