if 8; 2x; 2y form an arithmetic sequence and 2x; 2y; 36 form a geometric sequence determine the values of x and y
I feel like I am going around in circles on this problem.
For the AP i have the following formulas:
(1) d=(2y-8)/2
(2) d=2x-8
(3.1) (2y-8)/2=2x-8
or
(3.2) (2y-8)/2-(2x-8)=8
For the GP:
(1) 4y^2=72x
I know i am so how meant to use simultaneous equations to solve this , but I just cant seem to see how?
Any help with this problem would be appreciated.
Many thanks
For the AP:
2 x = 8 + d
2 y = 2 x + d = 8 + d + d = 8 + 2 d
2 y = 8 + 2 d
For the GP:
2 y = 36 / q
2 x = 2 y / q = ( 36 / q ) / q = 36 / q ^ 2
So:
2 x = 2 x
8 + d = 36 / q ^ 2
2 y = 2 y
8 + 2 d = 36 / q
Now you must solve system:
8 + d = 36 / q ^ 2
8 + 2 d = 36 / q
8 + 2 d = 36 / q Subtract 8 to both sides
8 + 2 d - 8 = 36 / q - 8
2 d = 36 / q - 8 Divide both sides by 2
d = ( 36 / q ) / 2 - 8 / 2
d = ( 18 / q ) - 4
Replace this value in equation:
8 + d = 36 / q ^ 2
8 + 18 / q - 4 = 36 / q ^ 2
4 + 18 / q = 36 / q ^ 2 Multiply bothsides by q ^ 2
4 q ^ 2 + 18 q ^ 2 / q = 36 q ^ 2 / q ^ 2
4 q ^ 2 + 18 q = 36 Subtract 36 to both sides
4 q ^ 2 + 18 q - 36 = 36 - 36
4 q ^ 2 + 18 q - 36 = 0
Try to solve this equation.
The solutions are :
q = - 6 and q = 3 / 2
For q = - 6:
d = 18 / q - 4
d = 18 / - 6 - 4
d = - 3 - 4
d = - 7
For q = 3 / 2:
d = 18 / ( 3 / 2 ) - 4
d = 18 * 2 / 3 - 4
d = 36 / 3 - 4
d = 12 - 4
d = 8
For q = - 6 and d = - 7
2 x = 8 + d
2 x = 8 + ( - 7 )
2 x = 8 - 7
2 x = 1 Divide both sides by 2
x = 1 / 2
2 y = 8 + 2 d
2 y = 8 + 2 * ( - 7 )
2 y = 8 - 14
2 y = - 6 Divide both sides by 2
y = - 6 / 2
y = - 3
For q = 3 / 2 and d = 8
2 x = 8 + d
2 x = 8 + 8
2 x = 16 Divide both sides by 2
x = 16 / 2
x = 8
2 y = 8 + 2 d
2 y = 8 + 2 * 8
2 y = 8 + 16
2 y = 24 Divide both sides by 2
y = 24 / 2
y = 12
All this mean you have 2 set of solutions:
1)
q = - 6 , d = - 7 , x = 1 / 2 , y = - 3
2)
q = 3 / 2 , d = 8 , x = 8 , y = 12
1 solution:
AP:
8 , 8 + d , 8 + 2 d
8 , 8 + ( - 6 ) , 8 + 2 * ( - 6 )
8 , 8 - 6 , 8 - 12
8 , 2 , - 4
GP:
2 x , 2 y , 2 y * q
2 * 1 / 2 , 2 * ( - 3 ) , 2 * ( - 3 ) * ( - 6 )
1 , - 6 , 2 * 18
1 , - 6 , 36
Or GP:
1 , 1 * q , 1 * q ^ 2
1 , 1 * ( - 6 ) , 1 * ( - 6 ) ^ 2
1 , - 6 , 1 * 36
1 , - 6 , 36
2 solution:
q = 3 / 2 , d = 8 , x = 8 , y = 12
AP:
8 , 8 + d , 8 + 2 d
8 , 8 + 8 , 8 + 2 * 8
8 , 16 , 8 + 16
8 , 16 , 24
GP:
2 x , 2 y , 2 y * q
2 * 8 , 2 * 12 , 2 * 12 * 3 / 2
16 , 24 , 24 * 3 / 2
16 , 24 , 72 / 2
16 , 24 , 36
Or GP:
16 , 16 * 3 / 2 , 16 * ( 3 / 2 ) ^ 2
16 , 48 / 2 , 16 * 9 / 4
16 , 24 , 144 / 4
16 , 24 , 36
I hope I was not too verbose.
Thanks so much Bosnian... It never crossed my mind to introduce a fourth variable. You are a rock star :D
To solve this problem, we can use the concept of arithmetic and geometric progression.
From the given information, we can write the following equations:
Arithmetic progression (AP):
1) (2x) - (8) = (8) - (2y)
Geometric progression (GP):
2) (36) / (2y) = (2y) / (2x)
Let's simplify these equations:
1) 2x - 8 = 8 - 2y
Rearranging, we get:
2x + 2y = 16
x + y = 8 ----(equation A)
2) 36 / (2y) = (2y) / (2x)
Simplifying, we get:
(y^2) = 36x ----(equation B)
Now, we have two equations with two unknowns (x and y). We can solve them simultaneously by substitution or elimination method.
Using substitution method:
We rearrange equation A as x = 8 - y.
Substituting this value in equation B, we get:
(y^2) = 36(8 - y)
Expanding, we have:
(y^2) = 288 - 36y
Rearranging, we get:
(y^2 + 36y - 288) = 0
Now, we can solve this quadratic equation to find the value(s) of y. Once we have the value(s) of y, we can substitute it back into equation A to find the corresponding value(s) of x.
Using quadratic formula, we have:
y = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = 1, b = 36, c = -288:
y = (-36 ± √(36^2 - 4(1)(-288))) / (2(1))
y = (-36 ± √(1296 + 1152)) / 2
y = (-36 ± √2448) / 2
Simplifying further:
y = (-36 ± √(3 * 2^5 * 17)) / 2
y = (-36 ± 2√(3 * 2^2 * 4 * 17)) / 2
y = (-36 ± 2 * 2 * √(3 * 17)) / 2
y = (-36 ± 4√(51)) / 2
y = -18 ± 2√(51)
So, we have two possible values for y:
y = -18 + 2√(51)
y = -18 - 2√(51)
Now, we substitute these values of y back into equation A to find the corresponding values of x:
For y = -18 + 2√(51):
x = 8 - (-18 + 2√(51))
x = 8 + 18 - 2√(51)
x = 26 - 2√(51)
For y = -18 - 2√(51):
x = 8 - (-18 - 2√(51))
x = 8 + 18 + 2√(51)
x = 26 + 2√(51)
Therefore, the possible values of x and y are:
(x, y) = (26 - 2√(51), -18 + 2√(51)) and (x, y) = (26 + 2√(51), -18 - 2√(51))
I hope this helps! Let me know if you have any more questions.