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x, y, z are 3 integers in a geometric sequence such that y − x is a perfect cube. Given log36x2+log6√y+log216y1/2z = 6. Find the value of x+y+z
- a)189
- b)190
- c)199
- d)201
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
x, y, z are 3 integers in a geometric sequence such that y − x i...
Possible values of (a, b) satisfying the equation:-
(1,36),(2,18),(3,12),(4,9),(9,4),(12,3),(18,2),(36,1)
Given y−x is a perfect cube
⇒ ab-a is perfect cube
⇒a(b−1) is perfect cube
Only possible when (a,b)=(9,4)
∴x=9,y=36,z=144
∴ x + y + z = 9 + 36 + 144 = 189
Most Upvoted Answer
x, y, z are 3 integers in a geometric sequence such that y − x i...
Let the common ratio be r. Then we know that y = xr and z = xr^2.
Since y is the geometric mean of x and z, we have:
y^2 = xz
Substituting in our expressions for y and z, we get:
(xr)^2 = x(xr^2)
Simplifying, we get:
r^2 = x
Substituting this into our expression for y, we get:
y = xr = x√x = x^(3/2)
So, x = y^(2/3)
Substituting this into our expression for z, we get:
z = xr^2 = y^(2/3) * r^2 = y^(2/3) * (y/x) = y^(5/3)
Therefore, x = y^(2/3), y = x^(3/2), and z = y^(5/3) form a geometric sequence.
Since y is the geometric mean of x and z, we have:
y^2 = xz
Substituting in our expressions for y and z, we get:
(xr)^2 = x(xr^2)
Simplifying, we get:
r^2 = x
Substituting this into our expression for y, we get:
y = xr = x√x = x^(3/2)
So, x = y^(2/3)
Substituting this into our expression for z, we get:
z = xr^2 = y^(2/3) * r^2 = y^(2/3) * (y/x) = y^(5/3)
Therefore, x = y^(2/3), y = x^(3/2), and z = y^(5/3) form a geometric sequence.
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