Geometric Progression What is the value of x and y if 2, x, y, 16 form a geometric progression ?
- print Print
- list Cite
Expert Answers
calendarEducator since 2010
write12,551 answers
starTop subjects are Math, Science, and Business
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
Related Questions
- Solve for x and y x^3-y^3=7 x^2+xy+y^2=7
- 2 Educator Answers
- What are x and y if 4^(x/y)*4^(y/x)=32 and log 3 (x-y)=1-log 3 (x+y) ?
- 2 Educator Answers
- Solve the system of equations algebraically x^2+y^2=100 x-y=2
- 1 Educator Answer
- What is the real solution of the system x^2+y^2=16, xy=3 ?
- 1 Educator Answer
- Equations .Solve for x and y if y=6/x and 2^(x+y)=32 .
- 1 Educator Answer
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, ....
Student Answers