SOLUTION: 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?

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Question 1106995: 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?
Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
If x is first term, and d the common difference, the sum to 16, when simplifed gives 2x%2B3d=8.

Squares-description gives %28x%2B3d%29%5E2=x%5E2%2B48.

Solve the system for d and x.
Substituting for 3d,
%28x%2B8-2x%29%5E2=x%5E2%2B48
%288-x%29%5E2=x%5E2%2B48
.
.
highlight%28x=1%29, the FIRST term of the progression.

Returning to 3d=8-2x
3d=8-2%2A1=6
3d=6
highlight%28d=2%29, common difference between successive terms.
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1,  3,  5,  7