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

The correct answer is (C):
Since x, y ,and z are in G.P. and x<y<z, let x = a, y=ar and z=ar2 , where a>0 and r>1.
It is also given that, 15x, 16y and 12z are in A.P.
Therefore, 2×16y=5x+12z
Substituting the values of x, y and z we get,
32ar = 5a+12ar2
⇒ 32r = 5 + 12r2
⇒ 12r2 - 32r + 5 = 0
On solving the above quadratic equation we get r=1/6 or 5/2.
Since r>1, therefore r=5/2.