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 Sequence Group V (Posted on 2011-03-31)
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.

 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.)
 Have to start somewhere. | Comment 1 of 4
If you start by writing everything in terms of A and B you get
C = AB/(2A-B)
D = A^2B/(2A-B)^2
E = AB^2/(2A-B)^2
I played with excel a bit but didn't discover much.

You can also use the last to solve for B, C and D in terms of A and E, but this requires the quadratic formula and leads to:
B = -2A(Eħ √(AE))/(E-A)

I haven't found C and D yet.
This actually shows some promise since it implies AE is a perfect square among other things.

 Posted by Jer on 2011-04-05 02:00:21

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