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Sequence Group V (Posted on 2011-03-31) Difficulty: 4 of 5
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)

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Possible approach and a suggested answer | Comment 2 of 4 |

I agree with Jer, as far as he goes:

C = (AB)/(2A-B) 
D = (A^2B)/(2A-B)^2 
E = (AB^2)/(2A-B)^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)/(2A-B-2b) 
B=(AB)/(2A-B-b) 

The smallest quintuplets I have found having a common difference (E-A) are:

A   b   B    C    c   D    d     E    E-A
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 2011-04-05 12:17:17
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