In a geometric progression, the fourth term exceeds the third term by $$24$$ and the sum of the second and third term is $$6$$. Then, the sum of the second, third and fourth terms is ------.

We can write the terms as $$6-a_3,a_3,a_3 +24$$

Since the terms are in GP, $$a_3^2 = a_4a_2$$

$$a_3^2 = (6-a_3)(a_3+24)$$

$$a_3^2 = (6-a_3)(a_3+24)$$

$$ 2a_3^2+18a_3-144 = 0$$

$$a_3^2+9a_3-72=0$$

Solving using the quadratic formula we get

$$ a_3 = \dfrac{-9\pm \sqrt{81+288}}{2} = \dfrac{-9\pm\sqrt{ 369}}{2}$$

$$a_3 = 5.1 \text{ or } -14.1$$

Thus, sum of the terms is 

$$a_3 + 30 = 35.1 \text{ or } 15.9$$