Question

等差数列｛an}中d=-2 a1=1 sn=-8 求n

Answer 1

sn=[2a1+(n-1)d]*n/2

-8=[2*1+(n-1)*(-2)]n/2

-16=[2+(n-1)*(-2)]n

-16=[2+2-2n]n

-16=(4-2n)n

-16=4n-2n^2

-8=2n-n^2

-8=2n-n^2

n^2-2n-8=0

(n-4)(n+2)=0

n=4或n=-2

所以n=4