Since x,y, and z are in G.P. and $x 0andr > 1$ 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=