S12=12a1+12×11d/2
=12(a3-2d)+66d
=12a3+42d
=42d+12×12
=42d+144>0
d>-24/7
S13=13a1+13×12d/2
=13(a3-2d)+78d
=13a3+52d
=52d+13×12
=52d+156
设等差数列前n项和Sn 已知a3=12 S12>0 S13
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