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Question 20 :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

- \\frac{1}{6})
- \\frac{3}{6})
- \\frac{3}{2})
- \\frac{5}{2})

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Since x, y, z are in GP , Common ratio (r) = \\frac{y}{x}) = \\frac{z}{y}) => xz = y^{2}

5x, 16y and 12z are in AP

16y – 5x = 12z- 16y

32y = 12z + 5x

Consider the terms in GP to be of the form \\frac{a}{r}), a, ar

Replacing the values in the equation,

32y = 12yr + \\frac{5y}{r})

32r = 12r^{2} - 5

12r^{2} – 32r + 5 = 0

12r^{2} – 30r – 2r+5 = 0

6r(2r -5) – 1(2r-5) = 0

(6r-1)(2r-5) = 0

r= \\frac{1}{6}) 𝑜𝑟 \\frac{5}{2})

Since its given that x < y < z r must be of a number > 1

Only **r = \\frac{5}{2})** satisfies that condition

The question is **"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" **

Choice D is the correct answer.

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