Question

已知等差数列{an}中,a3a7=-16,a4+a6=0,求{an}前n项和sn.要详解答案,我采纳!急用!

Answer 1

a3a7=(a1+2d)(a1+6d)=a1^2+8a1d+12d^2=-16

a4+a6=a1+3d+a1+5d=2a1+8d=0 a1=-4d

代入得

(-4d)^2+8*(-4d)d+12d^2=-16

d=2或d=-2

a1=-8或a1=8

Sn=na1+n(n-1)d/2

sn=n^2-9n或 sn=-n^2+9n