CAT 2018 Set-1 | Question: 67

Question

Let $x, y, z$ be three positive real numbers in a geometric progression such that $x < y < z$. If $5x$, $16y$, and $12z$ are in an arithmetic progression then the common ratio of the geometric progression is

• A. $3/6$
• B. $3/2$
• C. $5/2$
• D. $1/6$

Answer 1

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 $: \text {C}$