答：n=26
已知等差數(shù)列前4項(xiàng)和=21,后4項(xiàng)和=67,前n項(xiàng)和=286,即
A1+A2+A3+A4=21.(1)
A(n-3)+A(n-2)+A(n-1)+An=67.(2)
(1)+(2),得
(A1+A2+A3+A4)+[A(n-3)+A(n-2)+A(n-1)+An]=88
(A1+An)+[A2+A(n-1)]+[A3+A(n-2)]+[A4+A(n-3)]=88
在等差數(shù)列中
∵(A1+An)=[A2+A(n-1)]=[A3+A(n-2)]=[A4+A(n-3)]
∴4*(A1+An)=88
(A1+An)/2=11
Sn=[(A1+An)/2]*n
已知Sn=286,故
286=11n
n=26