Croatian Mathematical Competitions

Let the five numbers be $a_1, a_2, a_3, a_4, a_5$.

Given:
- $a_1 = -2$
- $a_5 = 6$
- $a_1, a_2, a_3, a_4$ are consecutive terms of an arithmetic progression (AP)
- $a_3, a_4, a_5$ are consecutive terms of a geometric progression (GP)

Let the common difference of the AP be $d$.
Then:
$$a_2 = a_1 + d = -2 + d$$
$$a_3 = a_1 + 2d = -2 + 2d$$
$$a_4 = a_1 + 3d = -2 + 3d$$

Let the common ratio of the GP be $r$.
Then:
$$a_4 = a_3 r$$
$$a_5 = a_3 r^2$$
But $a_5 = 6$, $a_3 = -2 + 2d$, $a_4 = -2 + 3d$.

From the GP:
$$a_4 = a_3 r \implies -2 + 3d = (-2 + 2d) r$$
$$a_5 = a_3 r^2 \implies 6 = (-2 + 2d) r^2$$

From the second equation:
$$6 = (-2 + 2d) r^2 \implies r^2 = \frac{6}{-2 + 2d}$$

From the first equation:
$$-2 + 3d = (-2 + 2d) r \implies r = \frac{-2 + 3d}{-2 + 2d}$$

But $r^2 = \left(\frac{-2 + 3d}{-2 + 2d}\right)^2 = \frac{6}{-2 + 2d}$

So:
$$\left(\frac{-2 + 3d}{-2 + 2d}\right)^2 = \frac{6}{-2 + 2d}$$
Multiply both sides by $(-2 + 2d)^2$:
$$(-2 + 3d)^2 = 6(-2 + 2d)$$
Expand:
$$(-2 + 3d)^2 = 4 - 12d + 9d^2$$
So:
4 - 12d + 9d^2 = 6(-2 + 2d) = -12 + 12d
Bring all terms to one side:
4 - 12d + 9d^2 + 12 - 12d = 0
$$$$
(4 + 12) + 9d^2 - 12d - 12d = 0
$$$$
16 + 9d^2 - 24d = 0
$$$$
9d^2 - 24d + 16 = 0
This is a quadratic equation in $d$. Solve for $d$:
d = $\text{24 24 2}$ - 4 $\cdot 9 \cdot$ 16}}{2 $\cdot$ 9}
$$$$
d = $\text{24 576}$ - 576}}{18}
$$$$
d = $\text{24}$ 0}{18} = $\frac{24}{18} = \frac{4}{3}$
$$So \$d = \frac{4}{3}\$. Now, compute the numbers:$$
a_1 = -2
$$$$
a_2 = -2 + $\frac{4}{3} = -\frac{2}{3}$
$$$$
a_3 = -2 + 2 $\cdot \frac{4}{3}$ = -2 + $\frac{8}{3} = \frac{2}{3}$
$$$$
a_4 = -2 + 3 $\cdot \frac{4}{3}$ = -2 + 4 = 2
$$$$
a_5 = 6
$$Check that \$a_3, a_4, a_5\$ are in geometric progression: Common ratio:$$
r = $\frac{a_4}{a_3} = \frac{2}{2/3}$ = 3
$$$$
Check $a_5 = a_3 r^2 = \frac{2}{3} \cdot 9 = 6$

Yes, it works.

Answer:

The numbers are $-2$, $-\frac{2}{3}$, $\frac{2}{3}$, $2$, $6$.