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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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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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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)
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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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This might be a dumb question but why can we assume the order of the GP starts with x, y, z?

We do know x<y<z but if the series starts with z first, then it could be in the order z, y, x with the common ratio being less than 1
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