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Let x, y, z be three positive real numbers in a geometric progression such that x < y < z. If 5x, 16y, and 12z are in an arithmetic progression then the common ratio of the geometric progression is
(2018)
- a)1/6
- b)3/2
- c)5/2
- d)3/6
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
Let x, y, z be three positive real numbers in a geometric progression ...
Since, 5x, 16y, 12z are in AP.
∴ 32y = 5x + 12z …(1)
∵ x, y, z are in GP
∴ y2 = xz ...(2)
Squaring both sides of (1), we get
1024y2 = 25x2 + 144z2 + 120xz
⇒ 1024xz = 25x2 + 144z2 + 120xz
⇒ 25x2+144z2 - 904xz = 0
⇒ 25x2 - 900xz - 4xz + 144 z2 = 0
⇒ 25x(x - 36z) - 4z(x - 36z) = 0
⇒ (25x - 4z) (x - 36z) = 0
∴
⇒
[r is the common ratio]
⇒
But
because x, y, z > 0 and x < y < z
∴ common ratio = 5 / 2
∴ 32y = 5x + 12z …(1)
∵ x, y, z are in GP
∴ y2 = xz ...(2)
Squaring both sides of (1), we get
1024y2 = 25x2 + 144z2 + 120xz
⇒ 1024xz = 25x2 + 144z2 + 120xz
⇒ 25x2+144z2 - 904xz = 0
⇒ 25x2 - 900xz - 4xz + 144 z2 = 0
⇒ 25x(x - 36z) - 4z(x - 36z) = 0
⇒ (25x - 4z) (x - 36z) = 0
∴
⇒
[r is the common ratio]⇒
But
because x, y, z > 0 and x < y < z∴ common ratio = 5 / 2
Most Upvoted Answer
Let x, y, z be three positive real numbers in a geometric progression ...
Let the common ratio of the geometric progression be r. We are given that x, y, z are in geometric progression, so we have the following equations:
y = rx
z = r^2x
We are also given that x + y + z = 96, so we can substitute the values of y and z in terms of x into this equation:
x + rx + r^2x = 96
x(1 + r + r^2) = 96
Since x is positive, we can divide both sides of the equation by (1 + r + r^2):
x = 96 / (1 + r + r^2)
We want to find the value of x. Let's substitute values of r and see what we get:
For r = 1, we have x = 96 / (1 + 1 + 1) = 96 / 3 = 32.
For r = 2, we have x = 96 / (1 + 2 + 4) = 96 / 7.
For r = 3, we have x = 96 / (1 + 3 + 9) = 96 / 13.
Out of these three values, x = 32 gives the largest value. Therefore, the maximum possible value of x is 32.
y = rx
z = r^2x
We are also given that x + y + z = 96, so we can substitute the values of y and z in terms of x into this equation:
x + rx + r^2x = 96
x(1 + r + r^2) = 96
Since x is positive, we can divide both sides of the equation by (1 + r + r^2):
x = 96 / (1 + r + r^2)
We want to find the value of x. Let's substitute values of r and see what we get:
For r = 1, we have x = 96 / (1 + 1 + 1) = 96 / 3 = 32.
For r = 2, we have x = 96 / (1 + 2 + 4) = 96 / 7.
For r = 3, we have x = 96 / (1 + 3 + 9) = 96 / 13.
Out of these three values, x = 32 gives the largest value. Therefore, the maximum possible value of x is 32.
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Let x, y, z be three positive real numbers in a geometric progression such that x < y < z. If 5x, 16y, and 12z are in an arithmetic progression then the common ratio of the geometric progression is(2018)a)1/6b)3/2c)5/2d)3/6Correct answer is option 'C'. Can you explain this answer?
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