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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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Some Thoughts had a go and reached nowhere | Comment 1 of 2

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
  Posted by Vishal Gupta on 2010-09-30 16:30:50

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