SOLUTION: find the three numbers in an arithmetic progression whose sum is 48 and the sum of their squares is 800

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Question 1017801: find the three numbers in an arithmetic progression whose sum is 48 and the sum of their squares is 800
Answer by ikleyn(52788) About Me  (Show Source):
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find the three numbers in an arithmetic progression whose sum is 48 and the sum of their squares is 800
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One can present the three consecutive terms of the AP as

x -d, x, x + d,

where x is the mid term and d is the common difference.

Then the sum of the tree terms is 3x, and you can easily find a from the equation

3x = 48,

which implies x = 48%2F3 = 16.

Now the sum of squares of the tree terms is

%2816-d%29%5E2+%2B+16%5E2+%2B+%2816%2Bd%29%5E2 = %28256+-+32d+%2B+d%5E2%29+%2B+256+%2B+%28256+%2B+32d+%2B+d%5E2%29 = 3%2A256+%2B+2d%5E2 = 768+%2B+2d%5E2.

It gives you an equation to find d:

768+%2B+2d%5E2 = 800  --->  2d%5E2 = 800+-+768 = 32  --->  d%5E2 = 32%2F2 = 16.

Hence, d = +/- 4.

It gives the AP terms as  12, 16 20,   or   20, 16, 12.

Answer. AP terms are  12, 16 20,   or   20, 16, 12.