等差数列{an}共有n项,前三项和为4,最后三项和为14,所有项和为54,则n=?

Answer 1

a(1)+a(2)+a(3)=4;
a(n)+a(n-1)+a(n-2)=14;
两式相加：因为a(1)+a(n)=a(2)+a(n-1)=a(3)+a(n-2)
所以有:3(a(1)+a(n))=18
=>a(1)+a(n)=6
=>S(n)=(a(1)+a(n))*n/2
=3n
=54
=>n=18