Four positive integers P
are such that P
(in this order) are in geometric sequence and Q
(in this order) are in harmonic sequence
Given that S
= 40, determine all possible quadruplet(s) (P
) that satisfy the given conditions.
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.