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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

  1. \\frac{1}{6})
  2. \\frac{3}{6})
  3. \\frac{3}{2})
  4. \\frac{5}{2})

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Explanatory Answer

Method of solving this CAT Question from Sequence and Series

Since x, y, z are in GP , Common ratio (r) = \\frac{y}{x}) = \\frac{z}{y}) => xz = y2
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 = 12r2 - 5

12r2 – 32r + 5 = 0
12r2 – 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"

Hence, the answer is \\frac{5}{2})

Choice D is the correct answer.

 

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