Question 1157937

The sum of the first three terms in a GP is 38. Their product is 1728. Find the values of the three terms.
My answer were 8 and 18. I want to confirm if it's correct and how will I find the values of the three terms with two solution of a and r
<pre>SUM of the 3 terms: {{{matrix(2,3, a[1] + a[1]r + a[1]r^2, "=", 38,
a[1](1 + r + r^2), "=", 38)}}}
                    {{{matrix(1,3, a[1], "=", 38/(1 + r + r^2))}}} ------- eq (i) 
Product of the 3 terms: {{{a[1](a[1]r)(a[1]r^2)}}}, and so: {{{matrix(1,3, a[1]^3r^3, "=", "1,728")}}} ======>  {{{matrix(3,3, a[1]^3, "=", "1,728"/r^3, a[1], "=", ("1,728"/r^3)^(1/3), a[1], "=", ("1,728"^(1/3))/(r^3)^(1/3))}}}  
                        {{{matrix(1,3, a[1], "=", 12/r)}}} --------- eq (ii)

We then get: {{{matrix(1,3, 38/(1 + r + r^2), "=", 12/r)}}}
             {{{matrix(1,3, 12(1 + r + r^2), "=", 38r)}}} ------ Cross-multiplying
             {{{matrix(1,3, 6(1 + r + r^2), "=", 19r)}}} ------- Factoring out GCF, 2
             {{{matrix(1,3, 6 + 6r + 6r^2, "=", 19r)}}}
             {{{matrix(1,3, 6r^2 + 6r  -  19r + 6, "=", 0)}}} 
             {{{matrix(1,3, 6r^2  -  13r + 6, "=", 0)}}}
             {{{matrix(1,3, 6r^2  -  9r  -  4r + 6, "=", 0)}}}
             3r(2r  -  3)  -  2(2r  -  3) = 0
             (3r  -  2)(2r  -  3) = 0
             3r  -  2 = 0		OR		2r  -  3 = 0
             3r = 2		        OR	        2r = 3
             {{{system(matrix(1,3, r, "=", 2/3), OR, matrix(1,3, r, "=", 3/2))}}}
{{{matrix(1,3, r, "=", 2/3)}}}
{{{matrix(1,3, a[1], "=", 12/r)}}}
{{{matrix(1,10, 1^(st), term, "or", a[1], "=", 12/(2/3), "=", 12 * (3/2), "=", 18)}}}
Therefore, if r, or common ratio = {{{2/3}}}, then {{{a[1]}}} or 1st term = 18
In this case, the 3 terms are: {{{highlight_green(matrix(1,13, "18,", "12,", "(18", "*", (2/3), ")", "and,", 8, "(", 18, "*", (2/3)^2, ")"))}}}
{{{matrix(1,3, r, "=", 3/2)}}}
{{{matrix(1,3, a[1], "=", 12/r)}}}
{{{matrix(1,10, 1^(st), term, "or", a[1], "=", 12/(3/2), "=", 12 * (2/3), "=", 8)}}}
Therefore, if r, or common ratio = {{{3/2}}}, then {{{a[1]}}} or 1st term = 8
In this case, the 3 terms are: {{{highlight_green(matrix(1,13, "8,", "12,", "(8", "*", (3/2), ")", "and,", 18, "(", 8, "*", (3/2)^2, ")"))}}}<pre>