$\text{ If three real numbers }x,y,z\text{ are in G.P. such that }x+y+z=42.\text{ If }x,\frac{5y}{4},z\text{ are in A.P., }$

$\text{then the largest possible value of x is }$

$\begin{array}{l}\text{ Let }x=a,y=ar,z=a{r}^2\text{ are in G.P }\\ \text{ Since }x+y+z=42\\ \Rightarrow a+ar+a{r}^2=42\dots \dots \dots \dots \dots ..\left(1\right)\end{array}$

$\begin{array}{l}\text{ Since }x,\frac{5y}{4},z\text{ are in A.P }\\ 2\left(\frac{5}{4}y\right)=x+z\\ \frac{5}{2}\cdot ar=a+a{r}^2\dots \dots \dots \dots ...\left(2\right)\end{array}$

Solving ,$ar=12$

Now, from (1)

$\begin{array}{l}a+12+a\left(\frac{144}{a^2}\right)=42\\ \Rightarrow a+\frac{144}{a}=30\\ \Rightarrow a^2-30a+144=0\\ \Rightarrow a^2-24a-6a+144=0\\ \Rightarrow a(a-24)-6(a-24)=0\\ \Rightarrow a=6\text{ or }24\\ \text{ Maximum of }a\text{ is }24\end{array}$