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(1)設等差數列的公差為d,由題意可得:a17=a1+16d,即-12=-60+16d,可解得d=3,∴an=-60+3(n-1)=3n-63.(2)由(1)可知an=3n-63,a30=27,所以數列前30項的和為:S30=30×(−60+27)2=-495
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