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:
Since x, y ,and z are in GP and x < y < z
Let x = a, y = ar and z = ar2, where a > 0 and r > 1
Also, 5x, 16y and 12z are in AP. So,
2 × 16y = 5x + 12z
Substituting the values of x, y and z,
32ar = 5a + 12ar2
⇒ 32r = 5 + 12r2
⇒ 12r2 − 32r + 5 = 0
On solving the quadratic equation,
r = 1/6 or 5/2
Since r > 1, therefore r = 5/2.
The correct option is C.