Three numbers are in GP such that their sum is 38 and their product is 1728. The greatest number among them is 

Let the numbers are $a,ar,a{r}^2$

Accoring to the question

$\begin{array}{r}a+ar+a{r}^2=38\\ a\left(1+r+r^2\right)=38\end{array}$   and   $a\cdot ar\cdot a{r}^2=1728$

$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{a}^3{r}^3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=1728\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}(ar{)}^3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=1728\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}ar\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=(1728{)}^{1/3}\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}ar\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=12\end{array}$

From Eqs. (i) and (ii), 

$\begin{array}{r}\frac{12}{r}\left(1+r+r^2\right)=38\\ \Rightarrow 12\left(1+r+r^2\right)=38r\\ \Rightarrow 6\left(1+r+r^2\right)=19r\\ \Rightarrow 6r^2+6r+6-19r=0\\ \Rightarrow 6r^2-13r+6=0\\ \Rightarrow 6r^2-9r-4r+6=0\\ \Rightarrow (2r-3)(3r-2)=0\\ \therefore r=\frac{3}{2},\frac{2}{3}\end{array}$

When $r=\frac{3}{2},\text{ then }a=8$

So, the series 8, 12, 18. 

$\text{ When }r=\frac{2}{3}\text{, then }a=18$

Series 18, 12, 8. 

Hence, greatest number is 18.