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Sequence Group IV (Posted on 2010-09-29) Difficulty: 3 of 5
Four positive integers P, Q, R and S with P < Q < R < S are such that P, Q and R (in this order) are in geometric sequence and Q, R and S (in this order) are in harmonic sequence.

Given that S - P = 40, determine all possible quadruplet(s) (P, Q, R, S) that satisfy the given conditions.

No Solution Yet Submitted by K Sengupta    
Rating: 4.0000 (1 votes)

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Hints/Tips reached somewhere (spoiler) Comment 2 of 2 |
Well, using Excel I tried a number of combinations and found one answer: (50,60,72,90).

Armed with an answer, it is not too hard to find where the last post went wrong.  Using Vishal's notation, the solution above corresponds to a = 50 and r = 6/5.

The sixth step is still correct:
ar^2 +r(a+40) -(2a+80) = 0

But it is not correction that step 8 
      9a^2 + 400a + 1600 >= 0 
implies that a lies between 40/9 and 40.  
9a^2 + 400a + 1600 is clearly >= 0 for any positive a.

And also, of course, r does not need to be integral if a is a multiple of a perfect square.  r needs to be integral if a = 5 or 6 or 7, but not if a = 8 or 9 or 12 or (more to the point) 50

Vishal is closer than I to determining if there are any more solutions, and I leave that task to Vishal or others.

  Posted by Steve Herman on 2010-09-30 22:46:51
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