s(n+1)=(n+1)(a1+a(n+1))/2,s(n+1)-s(n)=a(n+1)=(a1+(n+1)a(n+1)-n*a(n))/2，整理得a1+(n-1)a(n+1)-na(n)=0,又a1+na(n+2)-(n+1)a(n+1)=0;上两式相减得na(n+2)-2na(n+1)+na(n)=0,即a(n+2)-a(n+1)=a(n+1)-a(n).故为等差数列！
