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Geometric, Arithmetic and Harmonic Harness (Posted on 2014-10-19) Difficulty: 3 of 5
Each of A, B, C and D is a positive integer with A < B < C < D
having gcd(A, B, C, D) = 1 such that:

(i) A, B and C are in geometric sequence, and:
(ii) B, C and D are in arithmetic sequence, and:
(iii) A, B and D are in harmonic sequence.

Does there exist an infinite number of quadruplets satisfying the given conditions? Give reasons for your answer.

No Solution Yet Submitted by K Sengupta    
Rating: 3.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Analytical solution (spoiler) | Comment 1 of 3
My answer is NO, there are not an infinite number of quadruplets that satisfy.  In fact, I only find one.

Here is my analysis:

a) C = B^2/A, because of the geometric sequence

b) D = 2B^2/A - B, because of the arithmetic sequence

c) Then, because of the harmonic sequence,
   
   1/B - 1/A = (1-1/(2B^2/A - B)) - 1/B
   
   Combining and eliminating fractions gives
   (2A-B)(2B-A) = -A^2

then   3A^2 -5AB + 2B^2 = 0

then   (A-B)(3A - 2B) = 0
   
d) A = B is not an allowed solution, so the only solution is 3A = 2B,
   which gives rise to the lowest term integral solution of (4,6,9,12)

  Posted by Steve Herman on 2014-10-19 09:36:51
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