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 arithmetic sequence and Q, R and S (in this order) are in harmonic sequence.

Given that S - P = 19, determine all possible quadruplet(s) (P, Q, R, S) that satisfy the given conditions.

2Q=P+R --- (1)
2QS=RS+QR --- (2)
(1)*S-(2)
=>PS=RQ (Substitute Q=(P+R)/2  and S=19+P)
=>2Pē+38P=RP+Rē
P,Q,R are in A.P. So, R=P+2D (D is the common difference)
On simplification, we get
2Pē+38P=Pē+2DP+Pē+4DP+4Dē
=> 38P=6DP+4Dē
=>P(19-3D)=2Dē
=>P=2Dē/(19-3D)
P>0 => 19-3D>0 and D>0. So, D=1,2,3,4,5,6
=>Out of these 6 values, only D=6 satisfy the condition
that P is an integer.
So,P=72,Q=78,R=84,S=P+19=91.

Edited on July 22, 2008, 6:20 am

Posted by Praneeth on 2008-07-22 06:05:41