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

Answer 1

Answer:

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.

Answer 2

Answer:

I explained in above pic ,if any doubt msg me