Sn=na1+n(n-1)d/2
S3=3a1+3d=5 (1)
S9=9a1+36d=9 (2)
結合（1）（2）解,得：a1=2,d=-1/3
所以S6=6a1+15d=6*2+15*(-1/3)=7.
a10+a11+a12
=S12-S9
=12*a1+12*11/2(-1/3)-9
=-7.
等差數(shù)列{an}中,S3=5,S9=9,1.求S6
等差數(shù)列{an}中,S3=5,S9=9,1.求S6
2.求a10+a11+a12
2.求a10+a11+a12
數(shù)學人氣：898 ℃時間：2020-09-20 05:17:36
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