Question

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

Answer 1

The Correct Option is C

Let x, y, z be three positive real numbers in a geometric progression, so we have the relations \( y = xr \) and \( z = xr^2 \). Given that 5x, 16y, and 12z are in an arithmetic progression, the differences between successive terms are equal. Therefore:

16y - 5x = 12z - 16y

Substitute \( y = xr \) and \( z = xr^2 \): 

16(xr) - 5x = 12(xr^2) - 16(xr)

Simplify the expression:

16xr - 5x = 12xr^2 - 16xr

Combine like terms:

32xr = 12xr^2 + 5x

Divide the entire equation by \( x \) (since x > 0):

32r = 12r^2 + 5

Rearrange the equation to form a standard quadratic equation:

12r^2 - 32r + 5 = 0

Use the quadratic formula, \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a=12 \), \( b=-32 \), \( c=5 \).

Calculate the discriminant: \( b^2 - 4ac = (-32)^2 - 4 \cdot 12 \cdot 5 = 1024 - 240 = 784 \)

\( \sqrt{784} = 28 \)

Plug the values into the quadratic formula:

\( r = \frac{-(-32) \pm 28}{2 \cdot 12} = \frac{32 \pm 28}{24} \)

Calculate the possible values for r:

\( r_1 = \frac{32+28}{24} = \frac{60}{24} = \frac{5}{2} \)

\( r_2 = \frac{32-28}{24} = \frac{4}{24} = \frac{1}{6} \)

Since \( x < y < z \), \( r \) must be greater than 1, so the common ratio of the geometric progression is \( \frac{5}{2} \).