等差数列{an}首项及公差均是正整数,前n项和为Sn,且a1>1,a4>6,s3

S3=3a1+3d=3(a1+d)=3a2≤12 a2≤4a4=a2+2d≤2d+4,又a4>6,因此2d+4>6 d>1,d为正整数,d≥2.a1>1,a1为正整数,a1≥2.a2=a1+d≥2+2=4,又a2≤4,要两不等式同时成立,只有a2=4,a1=2,d=2.Sn=na1+n(n-1)d/2=2n+2n(n-1)/2=n...