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:

- 3/6
- 3/2
- 5/2
- 1/6

### Answer

Since x, y ,and z are in GP and x < y < z

Let x = a, y = ar and z = ar^{2}, 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 + 12ar^{2}

⇒ 32r = 5 + 12r^{2}

⇒ 12r^{2} − 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.**