Question 1150464
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            The "solution" by @josgarithmetic, giving the unique value of  {{{2/3}}}  to the common ratio,  is  UNCOMPLETED, 

            and,  therefore,  INCORRECT.


            I came to bring the correct solution.



<pre>
{{{a[3]}}} = 24

{{{a[5]}}} = {{{32/3}}}


{{{a[5]/a[3]}}} = r^2 = {{{((32/3))/24}}} = {{{32/72}}} = {{{4/9}}}, where r is the common ratio of the progression.


Therefore,  the common ratio r may have TWO possible values,  r = {{{2/3}}}  and  r = {{{-2/3}}}.

Therefore, looking for {{{a[10]}}}, we should consider TWO cases.


<U>Case 1</U>.   r = {{{2/3}}}.    Then  {{{a[10]}}} = {{{a[5]*r^5}}} = {{{(32/3)*(2^5/3^5)}}} = {{{2^10/3^6}}}.


<U>Case 2</U>.   r = - {{{2/3}}}.    Then  {{{a[10]}}} = {{{a[5]*r^5}}} = {{{(32/3)*(-2^5/3^5)}}} = - {{{2^10/3^6}}}.


<U>ANSWER</U>.  Under given conditions,  the 10-th term,  {{{a[10]}}},  may have one of the two values

         {{{a[10]}}} = {{{2^10/3^6}}}   or   {{{a[10]}}} = - {{{2^10/3^6}}}.
</pre>

Solved.