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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
  • a)
    3/6
  • b)
    1/6
  • c)
    5/2
  • d)
    3/2
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
Let x, y, z be three positive real numbers in a geometric progression ...
Let x = a, y = ar and z = ar2
It is given that, 5x, 16y and 12z are in AP.
so, 5x + 12z = 32y
On replacing the values of x, y and z, we get
5a + 12ar2 = 32ar
or, 12r2 – 32r + 5 = 0
On solving, r =5/2 or 1/6
For r = 1/6, x < y < z is not satisfied.
So, r=5/2
Hence, option C is the correct answer.
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Most Upvoted Answer
Let x, y, z be three positive real numbers in a geometric progression ...
Let x = a, y = ar and z = ar2
It is given that, 5x, 16y and 12z are in AP.
so, 5x + 12z = 32y
On replacing the values of x, y and z, we get
5a + 12ar2 = 32ar
or, 12r2 – 32r + 5 = 0
On solving, r =5/2 or 1/6
For r = 1/6, x < y < z is not satisfied.
So, r=5/2
Hence, option C is the correct answer.
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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 isa)3/6b)1/6c)5/2d)3/2Correct answer is option 'C'. Can you explain this answer?
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