Question

1.已知｛An｝是递增的等差数列,且前三项之和为12 ,钱3想之积48,求它的首项?

Answer 1

1.

a1+a2+a3=12,a1*a2*a3=48

因为2a2=a1+a3,所以3a2=12,a2=4.

a1+a3=8,a1*a3=48/4=12,因为是递增a1=2,a3=6

2.

a2+a5+a8=9,a2+a2+3d+a2+6d=9,3a2+9d=9

a3a5a7=-21,a3(a3+2d)(a3+4d)=-21

3a2+9d=3a3+6d=9,a3+2d=3,联立求得a3=-1,d=2

a3=a1+2d,a1=a3-2d=-5

an=-5+(n-1)2=2n-7