Asked by Alex Chien
The sum of four numbers in arithmetic progression is 16. The square of the last number is the square of the first number plus 48. What are the four numbers?
Answers
Answered by
bobpursley
A+A+d+A+2d+a+3d=16
4A+6d=16
2A+3d=8
(A+3d)^2=A^2+48
second equation...
A^2+6dA+9d^2=a^2+48
3d(2A+3d)=48
3d*8=48
d=2
then 2A+6=8
A=1
1,3,5,7
4A+6d=16
2A+3d=8
(A+3d)^2=A^2+48
second equation...
A^2+6dA+9d^2=a^2+48
3d(2A+3d)=48
3d*8=48
d=2
then 2A+6=8
A=1
1,3,5,7
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