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x, y, z are three positive numbers in a geometric progression such tha

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x, y, z are three positive numbers in a geometric progression such tha [#permalink] New post   12 May 2020, 21:49
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A
B
C
D
E

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x, y, z are three positive numbers in a geometric progression such that x < y < z, while 5x, 16y, and 12z are in an arithmetic progression. What is the common ratio of the geometric progression ?

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

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C
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Re: x, y, z are three positive numbers in a geometric progression such tha [#permalink] New post   Updated on: 13 May 2020, 00:43
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x, y, z are three positive numbers in a geometric progression such that x < y < z, while 5x, 16y, and 12z are in an arithmetic progression. What is the common ratio of the geometric progression ?

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

Let \(x = a, y = ar, z = ar^2\) and
Here r can't be less than 1 since x < y < z
So, A, D and E are eliminated

2*16y = 5x + 12z
\(32ar = 5a + 12ar^2\)
\(32r = 5 + 12r^2\)
\(12r^2 - 32r + 5 = 0\)

Let's pick C first.
r = \(\frac{5}{2}\),
\(12*\frac{25}{4} - 32*\frac{5}{2} + 5 = 0\)
\(75 - 16*5 + 5 = 0\)
LHS = RHS

Just to verify
If r = \(\frac{3}{2}\),
\(12*\frac{9}{4} - 32*\frac{3}{2} + 5 = 0\)
\(27 - 48 + 5 = 0\),
LHS ≠ RHS

Answer C.

Originally posted by unraveled on 12 May 2020, 22:33.
Last edited by unraveled on 13 May 2020, 00:43, edited 4 times in total.
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x, y, z are three positive numbers in a geometric progression such tha [#permalink] New post   13 May 2020, 00:09
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\(y= x*a\) , where a is the common ratio

\(z= x*a^2\)

5x, 16y, and 12z are in AP, then

\(2*16y = 5x +12z\)

\(32*xa = 5x + 12xa^2\)

\(12a^2-32a+5 = 0\)

\(12a^2-30a-2a+5=0\)

\(6a(2a-5) -1(2a-5) = 0\)

\(a= \frac{1}{6}\) or \(\frac{5}{2}\)

Since x<y<z, a must be greater than 1

\(a= \frac{5}{2}\)

C

OR

Since x<y<z, common ratio must be greater than 1

We can eliminate A, D and E

Option B- common ratio = 3/2

y=3x/2; z=9x/4

\(3x +12*\frac{9x}{4 }= 30x\) which is not equal to \(2*16*\frac{3x}{2}=48x\)

Eliminate B

C
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Re: x, y, z are three positive numbers in a geometric progression such tha [#permalink] New post   13 May 2020, 06:22
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Solution



    • Since x,y, and z are in G.P., let us assume that \(x = a , y = ar,\) and \(z = ar^2\)
      o Where a is a positive number and r is the common ratio.
    • Also, \(x < y < z ⟹ a < ar < ar^2\)
      o Therefore, \(r >1\)
    • Now, 5x, 16y, 12z are in A.P., therefore arithmetic mean of 5x and 12z must be equal to 16 y.
      i.e. \(\frac{5a + 12ar^2 }{ 2} = 16ar\)
      \(⟹ 5 + 12r^2 = 32r \)
      \(⟹12r^2 – 32r + 5 = 0\)
      \(⟹ 12r^2 – 30r – 2r + 5 = 0\)
      \(⟹(6r -1)(2r-5) = 0\)
      \(⟹ r = \frac{1}{6}\) or \(\frac{5}{2}\)
      Since, \(r > 1\), therefore \(r = \frac{5}{2}\)
Thus, the correct answer is Option C.
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Re: x, y, z are three positive numbers in a geometric progression such tha [#permalink] New post   13 May 2020, 06:46
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x, y, z are 3 positive numbers in a geometric progression such that x < y < z. So ratio must be greater than 1 (r>1).
Eliminate options (A), (D) and (E), because all of them have ratio less than 1 (r<1)

Suppose: y=x.r and z=x.r^2
Then, the following arithmetic progression 5x, 16y, and 12z can be substituted for 5x, 16x.r, and 12x.r^2.

16x.r - 5x = 12x.r^2 - 16x.r
32x.r - 5x = 12x.r^2

Since x is positive number, it's ok to divide by x
32r -5 = 12r^2
0 = 12r^2 -32r +5
0 = (2r-5)(6r-1)
r=5/2 (YES, r>1), or r=1/6 (NO, r<1)

FINAL ANSWER IS (C)

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x, y, z are three positive numbers in a geometric progression such tha [#permalink] New post   Updated on: 14 May 2020, 04:19
Quote:
x, y, z are three positive numbers in a geometric progression such that x < y < z, while 5x, 16y, and 12z are in an arithmetic progression. What is the common ratio of the geometric progression ?

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


com ratio: r=z/y=y/x, y^2=zx, y=sqrt(zx)
gprog: x=ar^0, y=ar^1, z=ar^2
aprog com dif: d=12z-16y=16-5x, y=(12z+5x)/32

equate/sub:
ar=12ar^2+5a/32,
12r^2+5-32r=0,
(2r-5)(6r-1)=0
r=1/6 or 5/2
common ratio >1=5/2

ans (C)

Originally posted by exc4libur on 13 May 2020, 18:23.
Last edited by exc4libur on 14 May 2020, 04:19, edited 1 time in total.
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Re: x, y, z are three positive numbers in a geometric progression such tha [#permalink] New post   13 May 2020, 21:22
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x, y, z are three positive numbers in a geometric progression such that x < y < z, while 5x, 16y, and 12z are in an arithmetic progression. What is the common ratio of the geometric progression ?

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

let common ration be r
y=xr,z=xr^2
32y=12z+5x
on subtituition
12r^2-32r+5=0
on solving
r=5/2 [r should be greater than 1]
Ans C
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Re: x, y, z are three positive numbers in a geometric progression such tha [#permalink] New post   09 Feb 2023, 22:42
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