Question 1151124
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Three positive numbers form an arithmetic progression; their sum is 18. If the first number is increased by 4, 
then the numbers will form a geometric progression. Find the original three numbers in arithmetic progression.
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<pre>
Since the three numbers form an AP with the sum 18, the middle terms is 18/3 = 6.


Let "d" be the common difference of this AP.


Then the three terms of the AP are  6-d, 6 and 6+d.


Then the three terms of the GP are  ((6-d)+4) = 10-d, 6 and 6+d.


Since the terms  10-d, 6 and 6+d  form a GP,


    {{{6/(10-d)}}} = {{{(6+d)/d}}}.


It is your equation to find "d". Cross multiply


    6d = (10-d)*(6+d)

    6d = 60 - 6d + 10d - d^2

    d^2 + 2d - 60 = 0

    {{{d[1,2]}}} = {{{(-2 +- sqrt(4 + 4*60))/2}}} = {{{(-2 +- sqrt(244))/2}}} = {{{-1 +- sqrt(61)}}}.



With the middle term 6,  <U>NEITHER</U>  value of  d = {{{-1 + sqrt(61)}}} = 6.81...  <U>NOR</U>   d = {{{-1 - sqrt(61)}}} = -8.81...

provides three positive terms of the AP.



<U>ANSWER</U>.  As the problem is worded, printed, posted and presented, <U>the solution DOES NOT exist</U>.
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