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.


Three numbers $a_1, a_2, a_3$





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

Let the three numbers are x, y and z.


(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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