填空题等差数列{an}的前n项和为Sn,若a1=20,S10=S15,则当n=________时,Sn最大.
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12或13解析分析:由S10=S15可得S15-S10=a11+a12+a13+a14+a15=0根据等差数列的性质可得,a13=0,结合a1=20>0?可得d<0??a12>0,a14<0从而可得可知S12=S13为Sn最大解答:∵S10=S15∴S15-S10=a11+a12+a13+a14+a15=0根据等差数列的性质可得,a13=0∵a1=20>0∴d<0??? a12>0,a14<0根据数列的和的性质可知S12=S13为Sn最大故