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 Going Greatest With Arithmetic, Geometric And Harmonic (Posted on 2008-03-22)
(A) Determine all possible non zero real P such that {P}, [P] and P are in arithmetic sequence.

(B) Determine all possible non zero real Q such that {Q}, [Q] and Q are in geometric sequence.

(C) Determine all possible non zero real R such that [R], {R} and R are in geometric sequence.

(D) Determine all possible non zero real S such that {S}, [S] and S are in harmonic sequence.

Note: [x] is defined as the greatest integer ≤ x and {x} = x - [x]

 See The Solution Submitted by K Sengupta Rating: 2.0000 (2 votes)

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 solutions | Comment 1 of 5

A) 1.5 (and almost: 2.99999999.....--i.e., writing 3 as this so its integer part would be 2)

B) phi = (sqrt(5)-1)/2

C) -1/phi^2

D) sqrt(2)/2 - 1 and sqrt(2)/2 + 1

A graphing calculator, or a computer program that produces graphs like one, is useful in finding between which integers the numbers lie, so as to know [x] and to formulate x - [x]. The graphing led to the pseudo-solution part of A. Plot y=x and also y = 2*int(x) - (x-int(x)) for the first example (A), and see where the broken curve crosses the straight line.

Once the integer portion is found so that it can be put explicitly in the equations, an arithmetic progression A,B,C is solved for C via 2B-A=C, with the appropriate substitution of, in the above case, R for C:

2(1)-(P-1)=P
2P=3
P=1.5

Similarly for geometric progression A,B,C, B^2 / A = C:

2(1^2)/(Q-1)=Q to get Q = phi

and with the harmonic progression, A,B,C, 2/B - 1/A = 1/C:

2/(-1) - 1/(S+1) = 1/S
and
2/1 - 1/(s-1) = 1/S.

 Posted by Charlie on 2008-03-22 15:17:21

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