Question 997811
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Let <B>x</B>&nbsp; be a middle number, &nbsp;then the first one is &nbsp;x-d &nbsp;and the third one is &nbsp;x+d, &nbsp;where &nbsp;d&nbsp; is the common difference.


The sum of three is &nbsp;12. &nbsp;It means that 


(x-d) + x + (x+d) = 12, &nbsp;&nbsp;&nbsp;&nbsp;or 


3x = 12, &nbsp;&nbsp;&nbsp;&nbsp;x = {{{12/3}}} = 4. 


The sum of squares is &nbsp;66. &nbsp;It means that


{{{(4-d)^2 + 4^2 + (4+d)^2}}} = {{{66}}}, &nbsp;&nbsp;&nbsp;&nbsp;or


{{{16 - 8d + d^2}}} + {{{16}}} + {{{16 + 8d + d^2}}} = {{{66}}},


{{{48}}} + {{{2d^2}}} = {{{66}}},


{{{2d^2}}} = {{{66-48}}},


{{{2d^2}}} = {{{18}}}, 


{{{d^2}}} = {{{18/2}}} = {{{9}}}, &nbsp;&nbsp;d = +/-{{{3}}}.


Thus our progression is &nbsp;&nbsp;1, 4, 7 &nbsp;&nbsp;or &nbsp;&nbsp;7, 4, 1.