Question 1017801
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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 = {{{48/3}}} = 16.

Now the sum of squares of the tree terms is

{{{(16-d)^2 + 16^2 + (16+d)^2}}} = {{{(256 - 32d + d^2) + 256 + (256 + 32d + d^2)}}} = {{{3*256 + 2d^2}}} = {{{768 + 2d^2}}}.

It gives you an equation to find d:

{{{768 + 2d^2}}} = {{{800}}}  --->  {{{2d^2}}} = {{{800 - 768}}} = 32  --->  {{{d^2}}} = {{{32/2}}} = 16.

Hence, d = +/- 4.

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

<U>Answer</U>. AP terms are  12, 16 20,   or   20, 16, 12.
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