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.

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)