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Three numbers form a geometric progression. If 8 is added to the second term, then these will constitute an arithmetic progression. If 64 is then added to the third term, the resulting numbers will form a geometric progression once again. Find the three given numbers.

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Three numbers $a_1, a_2, a_3$

$$\frac{a_3}{a_2}=\frac{a_2}{a_1}=\lambda$$

$$a_2+8-a_1=a_3-a_2-8=\omega$$

$$\frac{a_3+64}{a_2+8}=\frac{a_2+8}{a_1}=\mu$$

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It is simple to answer. There are three unknown and three equations:

Let the three numbers are x, y and z.

Then:

(1) y^2 = xz

(2) 2y + 16 = x + z

(3) (y+8)^2 = x(z+64)

On solving, we get:

(1) x = 4, y = 12 and z = 36

(2) x = 4/9, y = -20/9 and z = 100/9

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