S3=3×(a1+a3)/2=3(a1+d)=9a1=7, d=-5an=7-5(n-1)=12-5nSn=n(7+12-5n)/2=(19n-5n²)/2an24-10n<19n-5n²5n²-29n+24=(5n-24)(n-1)<01所以n=4为最大值
