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Let x, y, z be three positive real numbers in a geometric progression

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Let x, y, z be three positive real numbers in a geometric progression  [#permalink]

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New post Updated on: 28 Feb 2019, 23:43
1
00:00
A
B
C
D
E

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  45% (medium)

Question Stats:

38% (01:38) correct 63% (02:11) wrong based on 8 sessions

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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) 3/6
(B) 3/2
(C) 5/2
(D) 1/6
(E) 3/8

Originally posted by raghavrf on 28 Feb 2019, 23:27.
Last edited by raghavrf on 28 Feb 2019, 23:43, edited 1 time in total.
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Re: Let x, y, z be three positive real numbers in a geometric progression  [#permalink]

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New post 28 Feb 2019, 23:42
I am getting 2/5 and 1/6 as ans

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Re: Let x, y, z be three positive real numbers in a geometric progression  [#permalink]

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New post 28 Feb 2019, 23:54
5x, 16y, and 12z are in an arithmetic progression
this implies
2*16y = 12z + 5x
32y = 12z + 5x....(a)
x, y, z are three positive real numbers in a geometric progression
this implies
y = k*x.....(1)
z = k^2*x.....(2)
Therfore put (1) & (2) in (a)
32*kx = 12(k^2)x + 5x
32k = 12k^2 +5
or, 12k^2 -32k + 5 = 0
which gives k = 5/2 & k = 1/6
For k = 1/6, x < y < z is not satisfied.
Hence, k = 5/2
option(C)
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Re: Let x, y, z be three positive real numbers in a geometric progression   [#permalink] 28 Feb 2019, 23:54
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Let x, y, z be three positive real numbers in a geometric progression

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