CAT 2018 Slot 1 QA Question & Solution

AlgebraMedium

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

Options

$\frac{3}{6}$

$\frac{1}{6}$

$\frac{5}{2}$

$\frac{3}{2}$

Solution

Let x = $a$, y = $ar$ and z = $ar^2$
It is given that, 5x, 16y and 12z are in AP.
so, 5x + 12z = 32y
On replacing the values of x, y and z, we get
$5a + 12ar^2 = 32ar$
or, $12r^2 - 32r + 5$ = 0
On solving, $r$ = $\frac{5}{2}$ or $\frac{1}{6}$

For $r$ = $\frac{1}{6}$, x < y < z is not satisfied.

So, $r$ = $\frac{5}{2}$

Hence, option C is the correct answer.