CAT 2018 Quant Questions

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

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Correct Answer: 3

Let x = $a$, y = $ar$ and z = $ar^2$

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

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CAT 2018 Quant Questions
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