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10 Jul 2019, 01:01
Kudos
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C
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Question Stats:
50% (05:00) correct
50%
(03:23)
wrong
based on 2
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Official CAT 2018 Questions; Section: QA
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) 1/6
B) 3/2
C) 5/2
D) 3/6
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) 1/6
B) 3/2
C) 5/2
D) 3/6
06 Aug 2019, 13:39
Kudos
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Narenn
Official CAT 2018 Questions; Section: QA
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) 1/6
B) 3/2
C) 5/2
D) 3/6
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) 1/6
B) 3/2
C) 5/2
D) 3/6
x < y < z
—> y/x = r ; z/x = r^2
5x, 16y, and 12z are in AP
—> 2(16y) = 5x + 12z
—> 32y/x = 5 + 12z/x
—> 32r = 5 + 12r^2
—> 12r^2 - 32r + 5 = 0
—> 12r^2 - 30r - 2r + 5 = 0
—> (6r - 1)(2r - 5) = 0
—> r = 5/2 [r=1/6 is not possible as r>1 as per x<y<z]
IMO Option C
Pls Hit kudos if you like the solution
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