Given that, $x,y,z$ are in G.P.
Let $r$ be the common ratio of G.P.
Then, $ \frac {y}{x} = \frac{z}{y} = r \quad \longrightarrow (1) $
And, $5x,16y,12z$ are in A.P.
So,$16y-5x = 12z-16y$
$\Rightarrow 32y-5x=12z$
Divide both side by $y$ we get,
$12 \left(\frac{z}{y}\right) = 32-5 \left(\frac{x}{y}\right)$
$ \Rightarrow 12(r) = 32 – \frac {5}{r} \quad [ \because \text {from equation (1)}]$
$ \Rightarrow 12 r^{2} = 32r – 5$
$ \Rightarrow 12 r^{2} – 32r + 5 = 0$
$ \Rightarrow 12 r^{2} -30r -2r + 5 = 0$
$ \Rightarrow 6r (2r-5) – 1(2r-5) = 0$
$ \Rightarrow (2r-5) (6r-1) = 0$
$ \Rightarrow 2r-5 = 0 , 6r-1 = 0 $
$ \Rightarrow 2r=5 , 6r=1$
$ \Rightarrow r= \frac{5}{2} , r= \frac{1}{6}$
For $ r= \frac{5}{2} = \frac {y}{x} = \frac {z}{y}$
Here, $x:y=2:5, z:y=5:2$
We can write,
- $x:y = (2:5) \times 2 = 4 : {\color{Magenta}{10}}$
- $y:z = (2:5) \times 5 = {\color{Magenta}{10}} : 25$
$\therefore x:y:z = 4:10:25$
So, condition $x<y<z$ satisfied.
For $ r= \frac{1}{6} = \frac {y}{x} = \frac {z}{y}$
Here, $x:y=6:1, z:y=1:6$
We can write,
- $x:y = (6:1) \times 6 = 36 : {\color{Red}{6}}$
- $y:z = (6:1) \times 1 = {\color{Red}{6}} : 1 $
$\therefore x:y:z = 36:6:1$
So, condition $x<y<z$ doesn’t satisfied.
Therefore, $ \boxed {r= \frac{5}{2}}$
Correct Answer $: \text {C}$