Question

已知数列｛an｝是等差数列,其前n项和是Sn （1）若a1=1,an=-55,Sn=-405,求n及公差d

Answer 1

(1)sn=(an+a1)*n/2=(-55+1)*n/2=-405

n=15

sn=a1*n+n*(n-1)*d/2=1*15+15*(15-1)*d/2=-405

d=-4

(2)sn=a1*n+n*(n-1)*d/2=5/6*n+n*(n-1)*(-1/6)/2=5/2

n=

sn=(an+a1)*n/2=(an+5/6)*n/2=5/2

an=

(3)a3+a5+a7=a1+2d+a1+4d+a1+6d=3a1+12d=6

a2+a4+a6=a1+d+a1+3d+a1+5d=3a1+9d=-3

a1=-10 d=3