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# Geometric ProgressionWhat is the value of x and y if 2, x, y, 16 form a geometric...

gudeapp | Student, College Freshman | eNoter

Posted May 10, 2011 at 2:33 AM via web

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Geometric Progression

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

Tagged with algebra1, discussion

giorgiana1976 | College Teacher | Valedictorian

Posted May 10, 2011 at 10:20 AM (Answer #2)

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The geometric mean theorem of a g.p. states:

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 cancel each factor:

x = 0 and x^3 - 64 = 0

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

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

Let a = x and b = 4

x^3 - 64 = (x-4)(x^2 + 4x + 16)

(x-4)(x^2 + 4x + 16) = 0

We'll cancel each factor =>x - 4 = 0 => x = 4

We notice that x^2 + 4x + 16 > 0 for any real value of x.

If x = 4, we'll get y => 4^2 = 2y => y = 16/2 => y = 8

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

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted May 11, 2011 at 5:23 AM (Answer #3)

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Consecutive terms of a GP have a common ratio. if 2, x, y, 16 form a GP.

=> 16/y = x/2

=> x = 32/y

y/x = x/2

Substitute x = 32/y

=> y/(32/y) = (32/y)/2

=> 2y = (32/y)^2

=> 2y^3 = 32^2

=> y^3 = 32^2/2

=> y^3 = 2^(10 - 1)

=> y^3 =  2^9

=> y^3 = 8^3

=> y = 8

x = 32/y = 4

The value of x = 4 and y = 8

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