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Three Sequences (Posted on 2004-01-06) Difficulty: 3 of 5
A set of six positive integers contains an arithmetic sequence of four terms, a geometric sequence of four terms, and a harmonic sequence of four terms. What are the numbers in the set when the largest member of the set is a minimum?

Note: A harmonic sequence is a sequence whose reciprocals form an arithmetic sequence. ex: ({10, 12, 15, 20} is harmonic since {1/10, 1/12, 1/15, 1/20} is arithmetic)

See The Solution Submitted by Brian Smith    
Rating: 3.0000 (2 votes)

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Solution Puzzle Solution With Explanation Comment 9 of 9 |
(In reply to Answer by K Sengupta)

We know that the LCM of any four consecutive integers is the minimum if the said numbers are 1, 2, 3, 4. Since (1, 2, 3, 4) are also in arithmetic sequence, it follows that the smallest set of 4 positive integers in harmonic sequence is obtained by dividing LCM(1,2,3,4) = 12 in turn by 4, 3, 2, 1 giving 3, 4, 6, 12 as the smallest positive integral harmonic sequence consisting of precisely four  terms.

Now, by inspection, we observe that the smallest possible geometric sequence of four terms that will be obtained by appending at most  two numbers to 3, 4, 6, 12 is 3, 6, 12, 24 which is achieved by appending 24.

Thus five of the six required numbers are:
3, 4, 6, 12, 24.

Let the ith term of the given sequence (3,4,6,12,24) be denoted by p(i). Then, a sequence of four positive whole numbers in arithmetic sequence will be feasible upon insertion of an additional term whenever:

2[p(i) - (p(i-1] = p(i+1) - p(i)

By inspection, we observe that this is realised whenever:

(I) 6-4 = 2(4-3)
Accordingly, we insert 5 between 4 and 6, so that the six numbers are {3, 4, 5, 6, 12, 24}

(II) 12 - 6 = 2(6 -3)
Accordingly, we insert 9 between 6 and 12, so that the six numbers are {3, 4, 5, 6, 12, 24}

(III) 24 - 12 = 2(12-6)
Accordingly, we insert 18 between 12 and 24, so that the six numbers are {3, 4, 6, 12, 18, 24}


  Posted by K Sengupta on 2007-05-25 11:32:51
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