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Going Greatest With Arithmetic, Geometric And Harmonic (Posted on 2008-03-22) Difficulty: 2 of 5
(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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Solution Full Solution | Comment 4 of 5 |
Used in all cases: P = [P]+{P}

(A) 2*[P]={P}+P
=> 2*P-2*{P}={P}+P
=> P=3*{P}
=> [P]=2*{P} => 2*{p} must be an integer and {P}<1
This is possible only if {P}=0.5 or 0.
=> P=1.5 is the only possible non zero case possible.



(B) [Q]²={Q}Q => (Q/{Q})²-3Q/{Q}+1=0

Let Q/{Q}=x

=> x=(3+√5)/2 or (3-√5)/2

So, Q=(3+√5)/2*{Q} or (3-√5)/2*{Q}

=> [Q]²={Q}²*(3+√5)/2 or (3-√5)/2*{Q}²

=> [Q]= {Q}*(1+√5)/2  or {Q}*(√5-1)/2

If [Q]!=0, then |[Q]/{Q}| > 1
=> [Q]={Q}*(1+√5)/2
=> {Q}=(-1+√5)/2
=> Q = (1+√5)/2 = 1.618033988...

(C) {R}²=R*[R]

R=(3+√5)/2*[R] or (3-√5)/2*[R]

But if [R]>0, then 1 ≤|R/[R]| < 2

both of the equations dont satisfy this condition.

If [R]=0 => R=0.

If [R] ≤ -1, 0 < |R/[R]| ≤ 1
R=(3-√5)/2*[R]
=> {R}²=(3-√5)/2 => {R}=(√5-1)/2 => R=-(3-√5)/2 = -0.3819660

(D) 2/[S]=1/{S}+1/S

=> (S+{S})*(S-{S}) = 2*S*{S}

=> (S/{S})²-2*S/{S}-1=0

=> S = (√2 +1)*{S} or  (1-√2)*{S}

=> [S]=√2*{S} or -√2*{S}

=> 0≤{S}<1 => 0 ≤√2*{S} < √2 or -√2 ≤-√2*{S} < 0

[S] can take 1 or -1 as its values as it is an integer

=> {S}=1/√2

=> S = 1+1/√2 or -1+1/√2 = 1.707106 or -0.292893

Edited on March 24, 2008, 4:01 am
  Posted by Praneeth on 2008-03-24 03:39:36

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