數(shù)列共有2n＋1項(xiàng),首項(xiàng)為a1,公差為d,其奇數(shù)項(xiàng)有n＋1項(xiàng),偶數(shù)項(xiàng)有n項(xiàng),中間一項(xiàng)是第n＋1項(xiàng),則有
奇數(shù)項(xiàng)之和S1=(n+1)[a1+a1+2nd]/2=(n+1)(a1+nd)
偶數(shù)項(xiàng)之和S2=n[a1+d+(a1+d)+2(n-1)d]/2=n(a1+nd)
S1-S2=a1+nd=a(n+1)=96-80=16
所以中間項(xiàng)是16
因?yàn)閍(1)+a(2n+1)=2*a(n+1),a(3)+a(2n-1)=2*a(n+1),...
a(1)+a(3)+...+a(2n+1)=(n+1)*a(n+1)
同樣有a(2)+a(4)+...+a(2n)=n*a(n+1)
所以：(n+1)/n=96/80=6/5
n=5