Question

已知四个数依次成等差数列,且四个数的平方和为94,首尾两数之积比中间两数之积少18,求此等差数列.

Answer 1

设四个数是a-d,a,a+d,a+2d

所以(a-d)^2+a^2+(a+d)^2+(a+2d)^2=94

4a^2+4ad+6d^2=94

(a-d)(a+2d)+18=a(a+d)

a^2+ad-2d^2+18=a^2+ad

d^2=9

d=3d=-3

4a^2+4ad+6d^2=94

若d=3,a^2+3a-10=0

a=-5,a=2

若d=-3,a^2-3a-10=0

a=5,a=-2

所以是-8,-5,-2,1

或-1,2,5,8

或8,5,2,-1

或1,-2,-5,-8