Question

已知等差数列{an}中,满足a3*a7=-16,a4+a6=0,求数列{an}的前n项和Sn

Answer 1

a3+a7=a4+a6=0

a7=-a3

a3a7=-16

所以-a3?=-16

a3=±4

a3=4,a7=-4

4d=a7-a3=-8

d=-2

a1=a3-2d=8

an=-2n+10Sn=(a1+an)n/2=n?+9n

a3=-4,a7=4

4d=a7-a3=8

d=2

a1=a3-2d=-8

an=2n-10Sn=(a1+an)n/2=-n?-9n

所以Sn=n?+9n或Sn=-n?-9n