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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: $$a_4=a_{3}r⟹-2+3d=(-2+2d)r$$ $$a_5=a_3{r}^2⟹6=(-2+2d)r^2$$

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

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

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 $$Bringalltermstooneside:$$ 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 = \frac{24 \pm \sqrt{24^2 - 4 \cdot 9 \cdot 16}}{2 \cdot 9} d = \frac{24 \pm \sqrt{576 - 576}}{18} d = \frac{24 \pm 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$a_5 = a_3 r^2 = \frac{2}{3} \cdot 9 = 6$<br/><br/>Yes,itworks.<br/><br/>\ast \ast Answer:\ast \ast <br/><br/>Thenumbersare$-2$,$-\frac{2}{3}$,$\frac{2}{3}$,$2$,$6$.