If three real numbers x,y,z are in G.P. such that x+y+z=42. If x,5y4,z are in A.P., then  the largest possible value of x is

Let x=a,y=ar,z=ar2 are in G.P  Since x+y+z=42⇒a+ar+ar2=42……………..(1) Since x,5y4,z are in A.P 254y=x+z52⋅ar=a+ar2…………...(2)Solving ,ar=12Now, from (1)a+12+a144a2=42⇒a+144a=30⇒a2−30a+144=0⇒a2−24a−6a+144=0⇒a(a−24)−6(a−24)=0⇒a=6 or 24 Maximum of a is 24