Question

已知｛An｝为等差数列Sn为其前n项和若A1+A3+A5=-12,S7+S9=S11则求数列｛Sn/n｝的前8项的和是多少

Answer 1

设数列的第一项为a1，公差为d
则a3=a1+(3-1)d=a1+2d，a5=a1+(5-1)d=a1+4d
a1+a3+a5=a1+a1+2d+a1+4d=3a1+6d=-12 (*)
S7=(a1+a7)*7/2=(a1+a1+(7-1)d)*7/2=7a1+21d
S9=(a1+a9)*9/2=(a1+a1+(9-1)d)*9/2=9a1+36d
S11=(a1+a11)*11/2=(a1+a1+(11-1)d)*11/2=11a1+55d
S7+S9=S11，即7a1+21d+9a1+36d=11a1+55d，解得a1=-2d/5
将此关系代回(*)式，解得d=-5/2，则a1=-2d/5=-2/5*(-5/2)=1
Sn/n=(a1+an)*n/2/n=(a1+a1+(n-1)d)/2=a1-d/2+nd/2
所以Sn/n的前8项之和为：8*(a1-d/2)+(1+8)*8/2*d/2=8a1+14d=8*1+14*(-5/2)=-27
答案也就是-27