What is the value of x and y if 2, x, y, 16 form a geometric progression.

We'll use the theorem of geometric mean of a
g.p.:

x^2 = 2y (1)

y^2 = 16x
(2)

We'll raise to square (1):

x^4 =
4y^2

We'll divide by 4 both sides:

y^2
= x^4/4 (3)

We'll substitute (3) in
(2):

x^4/4 = 16x

We'll cross multiply
and we'll get:

x^4 = 4*16x

We'll
subtract 64 both sides:

x^4 -  64x =
0

We'll factorize by x:

x(x^3 - 64) =
0

We'll put each factor as 0:

x =
0

x^3 - 64 = 0

We'll re-write the
difference of cubes, using the formula:

a^3 - b^3 = (a-b)(a^2 + ab +
b^2)

We'll put a = x and b = 4

x^3 - 64
= (x-4)(x^2 + 4x + 16)

(x-4)(x^2 + 4x + 16) =
0

x - 4 = 0

x =
4

x^2 + 4x + 16 > 0 for any value of
x.

For x = 4, we'll get y:

4^2 =
2y

y = 16/2

y =
8

So, for x = 4 and y = 8, the terms of the geometric
series, whose common ratio is r  =2, are: 2 , 4 , 8 , 16,
....