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If a, b, x, y, z are positive numbers such that a, x, b are in A.P. and a, y, b are in G.P. and z=(2ab)/(a+b)then
- a)xyz are in G.P.
- b)x ≥ y ≥ z
- c)Both
- d)None
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
If a, b, x, y, z are positive numbers such that a, x, b are in A.P. an...
Since a, x, b are in A.P., we know that x = (a+b)/2.
Since a, y, b are in G.P., we know that y = sqrt(ab).
Substituting these values into z=(2ab)/(a b), we get:
z = (2ab)/(a*b)
z = 2(a*b)/(a*b)
z = 2
Therefore, we have:
x = (a+b)/2
y = sqrt(ab)
z = 2
a) To check if x, y, z are in G.P., we need to see if y^2 = xz.
y^2 = (sqrt(ab))^2
y^2 = ab
xz = (a+b)/2 * 2
xz = a+b
Since y^2 = ab and xz = a+b, we have:
y^2 = xz
Therefore, x, y, z are in G.P.
b) To find the value of a/b in terms of x and y, we can use the formula for the G.P.:
y^2 = ab
Substituting x = (a+b)/2, we get:
y^2 = x(b-x/2)
y^2 = xb - x^2/2
ab = xb - x^2/2
a = (xb - x^2/2)/b
a/b = x - x^2/(2b)
Therefore, the value of a/b in terms of x and y is a/b = x - x^2/(2b).
Since a, y, b are in G.P., we know that y = sqrt(ab).
Substituting these values into z=(2ab)/(a b), we get:
z = (2ab)/(a*b)
z = 2(a*b)/(a*b)
z = 2
Therefore, we have:
x = (a+b)/2
y = sqrt(ab)
z = 2
a) To check if x, y, z are in G.P., we need to see if y^2 = xz.
y^2 = (sqrt(ab))^2
y^2 = ab
xz = (a+b)/2 * 2
xz = a+b
Since y^2 = ab and xz = a+b, we have:
y^2 = xz
Therefore, x, y, z are in G.P.
b) To find the value of a/b in terms of x and y, we can use the formula for the G.P.:
y^2 = ab
Substituting x = (a+b)/2, we get:
y^2 = x(b-x/2)
y^2 = xb - x^2/2
ab = xb - x^2/2
a = (xb - x^2/2)/b
a/b = x - x^2/(2b)
Therefore, the value of a/b in terms of x and y is a/b = x - x^2/(2b).
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Community Answer
If a, b, x, y, z are positive numbers such that a, x, b are in A.P. an...
Hence,
a, x, b are in A.P.
Then,
x-a = b-x
... a+b = 2x
a, y, b are in G.P.
Then,
y/a = b/y
... y^2 = ab
and given, z = 2ab/(a+b)
z = 2y^2/2x
zx = y^2
zx = y*y
...z/y = y/x
Hence, x, y, z are in G.P and there is also possibility for x≥y≥z .Hence, the correct answer is option 'C' .
a, x, b are in A.P.
Then,
x-a = b-x
... a+b = 2x
a, y, b are in G.P.
Then,
y/a = b/y
... y^2 = ab
and given, z = 2ab/(a+b)
z = 2y^2/2x
zx = y^2
zx = y*y
...z/y = y/x
Hence, x, y, z are in G.P and there is also possibility for x≥y≥z .Hence, the correct answer is option 'C' .
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