Question

已知数列{An}是等差数列,前四项和为21,末四项和为67,且前n项和为286则n=_____.

Answer 1

a1+ a2 + a3 + a4 =21①

a(n-3)+a(n-2)+a(n-1)+an=67

即an+a(n-1)+a(n-2)+a(n-3)=67②

两式相加,得

4(a1+an)=21+67=88

a1+an=22

又Sn=(a1+an)n/2

所以n=26