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): 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(52790) (Show Source):
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