Five positive integers
A,
B,
C,
D and
E, with
A <
B <
C <
D <
E, are such that:
(i)
A,
B and
C (in this order) are in
harmonic sequence, and:
(ii)
B,
C and
D (in this order) are in geometric sequence, and:
(iii)
C,
D and
E (in this order) are in arithmetic sequence.
Determine the minimum value of (
E
A) such that there are precisely three quintuplets (
A,
B,
C,
D,
E) that satisfy all the given conditions.
I agree with Jer, as far as he goes:
C = (AB)/(2AB)
D = (A^2B)/(2AB)^2
E = (AB^2)/(2AB)^2
Let b be the difference constant in the harmonic series; since we know C, A and B can then be represented as
A=(AB)/(2AB2b)
B=(AB)/(2ABb)
The smallest quintuplets I have found having a common difference (EA) are:
A b B C c D d E EA
6 5 11 66 6 396 330 726 720
30 20 50 150 3 450 300 750 720
90 45 135 270 2 540 270 810 720
(b,c,d, are the difference constants) but I lack the facilities for an exhaustive search, and this result involves some 'lateral thinking' about the sort of results that might work.

Posted by broll
on 20110405 12:17:17 