2, x, y, 16 form a geometric progression. Therefore the ratio of consecutive terms is the same.

x/2 = y/x … (1)

y/x = 16/y … (2)

x/2 = y/x

=> x^2 = 2y

=> y = x^2 /2

y/x = 16/y

=> y^2 = 16x

substitute y = x^2 /2

=> x^4 / 4 = 16x

=> x^3 = 16*4

=> x = 4

y = x^2 /2 = 16/2 = 8

**Therefore we find that x = 4 and y = 8.**

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, ....**

2,x,y and 6 are in GP.

To determine x and y.

We use the property that the consecutive terms bear the same ratio to determine x and y.

So x/2 = y/x = 16/y.

Fro the 1st and last we get: x/2 = 16/y . So xy = 32..(1)

From the first and 2nd ratio, we get: x/2 = y/x . So y = x^2/2....(2).

So we substitute y = x^2/2 in xy = 32 and we get:

x(x^2/2) = 32.

So x^3 = 64. Or x = (64)^(1/3) = 4.

Therefore y = x^2/2 we get , y = 8.

Threfore x= 4 and y = 8.