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.
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
2/R= 1/Q +1/S
2/ar^2 = 1/ar + 1/S
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