Question 1157937
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Let "a" be the first term and "r" be the common ratio.


Then from the condition, we have these two equations

    a + ar + ar^2  =   38,      (1)

    a*(ar*)*(ar^2) = 1728.      (2)


From equation (2),  a^3*r^3 = 1728,  or  (ar)^3 = 1728,   which implies


    ar = {{{root(3,1728)}}} = 12;          (3)    

hence,  

    r  = {{{12/a}}}.                   (4)


Now, in equation (1) replace the term  ar  by 12, based on (3).  You will get

    a + 12 + ar^2 =  38,   which implies

    a + ar^2 = 26.              (5)


Next, substitute  r = {{{12/a}}}  into equation (5), replacing "r" there.  You will get

    {{{a}}} + {{{a*(144/a^2)}}} = 26,   or

    {{{a}}} + {{{144/a}}} = 26.


Multiply by "a" both sides and simplify

    a^2 - 26a + 144 = 0,

    {{{(a-13)^2}}} - 169 + 144 = 0

    {{{(a-13)^2}}} = 25

    a - 13 = +/- {{{sqrt(25)}}} = +/- 5.


Thus two solutions for "a" are  a = 13 + 5 = 18  or  a = 13 - 5 = 8.


If  a =  8, then from (4)  r = {{{12/8}}} = {{{3/2}}}.

If  a = 18, then from (4)  r = {{{12/18}}} = {{{2/3}}}.
    


In the first case, if a = 8,  then the three terms are  8, {{{8*(3/2)}}} = 12  and  {{{8*(3/2)^2}}} = 18.

    In this case, the sum of terms is  8 + 12 + 18 = 38, so this solution does work.



In the second case, if a = 18,  then the three terms are  18, {{{18*(2/3)}}} = 12  and  {{{18*(2/3)^2}}} = 8.

    In this case, the sum of terms is  18 + 12 + 8 = 38, so this solution does work, too.



<U>ANSWER</U>.  The problem has two solution:  

         a)  first term is 18;  the common difference is {{{2/3}}}  and the progression is  18, 12, 8.

         b)  first term is  8;  the common difference is {{{3/2}}}  and the progression is   8, 12, 18.
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Solved.