SOLUTION: If 4/3 ,M, 1, N is form GP, what is the product of M and N

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Question 1204480: If 4/3 ,M, 1, N is form GP, what is the product of M and N
Found 3 solutions by josgarithmetic, greenestamps, ikleyn:
Answer by josgarithmetic(39618)   (Show Source): You can put this solution on YOUR website!
If they are in geometric progression then you have these:


The very simplest of algebra gives .

This should still be checked with the first given term.





-----but this seems completely unrelated to the actual question for the problem.



(second part was rechecked)

Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!


Tutor @josgarithmetic has the right answer immediately... but in trying to check her result she does some sloppy algebra, arriving at a contradiction.

In a geometric progression, the square of any term is equal to the product of the two terms on either side of it. Using that fact, we get the answer immediately:

1^2 = (M)(N)
MN = 1

ANSWER: MN = 1

The problem doesn't ask us to find M and N; but we can easily.

M^2 = (4/3)(1)
M^2 = 4/3
M = 2/sqrt(3)

That gives us sqrt(3)/2 as the common ratio; and that gives us N = sqrt(3)/2.

The terms of the GP are 4/3, M=2/sqrt(3), 1, and N=sqrt(3)/2.


Answer by ikleyn(52792)   (Show Source): You can put this solution on YOUR website!
.

Any geometric progression has this characteristics property



   for any three consecutive terms  ,   and   

   the square of a middle term is equal to the product of its neighbors



For the proof, see the lesson

    One characteristic property of geometric progressions

in this site.


From this property, for your progression  M*N =  = 1.    ANSWER



Notice that it is true independently of the given fact 
that 4/3 is the first term of this progression.


In other words, this fact is excessive and unnecessary information for the problem.

Solved and explained.




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