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.

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.