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# For a list of positive integers, from the third terms, each

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Manager
Joined: 19 Aug 2006
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For a list of positive integers, from the third terms, each [#permalink]  04 Jan 2007, 22:40
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For a list of positive integers, from the third terms, each term is the sum of two previous terms, if the sum is odd; otherwise, the term is half of the sum. If the fourth term is 7, and the fifth term is 5, what is the first number?

Sorry friends dont have OA for this.
SVP
Joined: 01 May 2006
Posts: 1798
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So, we are left with these sequences' formula:
o a(n) = a(n-1) + a(n-2) if a(n-1) + a(n-2) = 2k+1 where k and n are integers
or
o a(n) = (a(n-1) + a(n-2))/2 if a(n-1) + a(n-2) = 2k where k and n are integers

Let see case by case what happens.

The third term
Hyp 1 :
> a(5) = a(4) + a(3)
<=> 5 = 7 + a(3)
<=> a(3) = -2 : But, the problem speaks a list of positive integers. So that Hyp 1 does not match.

Hyp 2 :
> a(5) = (a(4) + a(3)) / 2
<=> 10 = 7 + a(3)
<=> a(3) = 3 >>> Ok

The second term
Hyp 1 :
> a(4) = a(3) + a(2)
<=> 7 = 3 + a(2)
<=> a(2) = 4 >>> seems ok for the moment

Hyp 2 :
> a(4) = (a(3) + a(2)) / 2
<=> 14 = 3 + a(2)
<=> a(2) = 11 >>> seems ok for the moment

The first term
Here, we still have 2 possible values for a(2). We will check the validity in each case.
Hyp 1 : a(2) = 4
> a(3) = a(2) + a(1)
<=> 3 = 4 + a(1)
<=> a(1) = -1 >>> Not matching : the list must be exclusively composed of positive terms.

Hyp 2 : a(2) = 4
> a(3) = (a(2) + a(1)) / 2
<=> 6 = 4 + a(1)
<=> a(1) = 2 >>> Ok.

Hyp 3 : a(2) = 11
> a(3) = a(2) + a(1)
<=> 3 = 11 + a(1)
<=> a(1) = -8 >>> Not matching : the list must be exclusively composed of positive terms.

Hyp 4 : a(2) = 11
> a(3) = (a(2) + a(1)) / 2
<=> 6 = 11 + a(1)
<=> a(1) = -5 >>> Not matching : the list must be exclusively composed of positive terms.

Finally, we have found a(1) = 2.
Senior Manager
Joined: 24 Nov 2006
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LetÂ´s assume the sequence goes a1, a2, a3, 7, 5...

For two contiguous terms x and y:
If x+y even => next term = (x+y)/2.
If x+y odd => next term = x+y.

By simple inspection we notice that, for a list of +ve integers, a3 + 7 had to be even (otherwise a3 = -2, not possible given that all as must be +ve), thus a3 + 7 = 10 => a3 = 3.

a2 + 3 = 7 OR a2 + 3 = 7*2 = 14.
a2 = 4 OR a2 = 11.

a1 + a2 = 3 OR a1 + a2 = 3*2 = 6

For a2 = 4: a1 = -1 (not possible) OR a1 = 2 (possible).

For a2 = 11: a1 = -8 (not possible) OR a1 = -5 (not possible).

Only possible solution is a1 = 2.
Director
Joined: 26 Feb 2006
Posts: 905
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Re: PS [ Sequence ] [#permalink]  05 Jan 2007, 21:23
johnycute wrote:
For a list of positive integers, from the third terms, each term is the sum of two previous terms, if the sum is odd; otherwise, the term is half of the sum. If the fourth term is 7, and the fifth term is 5, what is the first number?

Sorry friends dont have OA for this.

since there are no -ves, the fifth term = 1/2 (3rd term + 4th term) = 5.
so 3rd term = 10 - 7 = 3

since 4th term = 7, which be could be either (2nd term + 3rd term) or 1/2 of (2nd term + 3rd term), the secon term could be 4 or 11 but 3rd term is 4 so 2nd term cannot be 11. therefore, 2nd term is 4.

third term is 3, which is equal to (1st term + 2nd term)/2. therefore, the first term should be 2.
Manager
Joined: 03 Dec 2006
Posts: 77
Location: London
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We have x1,x2,x3,7,5 We need to find x1

We know that the sum of x3 and 7 has been divided by 2 (otherwise x5 would be more than 7).

Therefore x3 + 7 =10 --> x3 = 3

so we get x1,x2,3,7,5

We know that x2 + 3 = 7 ( because 7 is odd and is not less than 4)
--> x2 = 4

so we get x1,4,3,7,5

We know that the sum of x1 and 4 has been divided by 2 (because 3 is less than 4)

Therefore x1 + 4 = 6 --> x1 = 2
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