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.

Spent a good amount of time on this. And somehow i ended up proving this was not possible. Probably an error on my part. Please do point out where i am making a mistake.

Terms in geometric sequence P, Q, R can be written as a, ar, ar^2

Harmonic progression

2/R= 1/Q +1/S

2/ar^2 = 1/ar + 1/S

1/S= (2-r)/(ar^2)

S= ar^2/(2-r)

S-P = 40

(ar^2 - 2a +ar) /(2-r) = 40

ar^2 +r(a+40) -(2a+80) = 0

Quadratic in r so discriminant >= 0

a^2+80a+1600 + 8a^2 +320a >= 0

9a^2 + 400a + 1600 >= 0

a (and hence P) lies between **40/9 and 40**.

min value of P = 5.

**r has to be integral** for 5r and 5r^2 to be integral.

r=2 can never be answer as then 2/R = 1/Q + 1/S will mean

2/4P = 1/2P + 1/S or S=infinity.

For P = 5, r=3-> Q=15, R=45, S=45 does not satisfy.

For P = 6, r=3 -> Q=18, R=54, S=46 which does not satisfy. (R has to be less than S)

For any higher values of P. it can never be true.

as P>=6 and r>=3 will make R greater than S.

So no solution should exist.

*Edited on ***September 30, 2010, 4:33 pm**