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

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

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07 Aug 2006, 17:44
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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 fourth term is 7, and fifth term is 5, what is the first term?

A. 2
B. 3
C. 4
D. 5
E. 7
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VP
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07 Aug 2006, 17:55
A. 2

Given S5 = 5, S4 = 7 => S3= 5x2-7 = 3

S3=3, S4=7 => S2 = 4
S2=4, S3=3, => S1= 2x3-4 = 2

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SVP
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08 Aug 2006, 00:50
A 2

_ _ _ 7 5

7+x = 5

X will be a negative number. Not possible
Hence 5 = (7+x)/2
X = 3

Hence the third term is 3
Since the third term is 3, the second term would be 4 and the first term would be 2

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CEO
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08 Aug 2006, 01:03
A

xyz75. x, y and z are positive

z must be 3 or -2. can not be -2. So z = 3
Now xy375
y must be 4 or 11.
When x4375:
x must be 2 or -1. Can not be -1. So its 2.
When x,11,375:
x must be -8 or -5. Both are -ve so not possible.

Finally x = 2
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Manager
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11 Aug 2006, 22:06
jaynayak wrote:
A 2

_ _ _ 7 5

7+x = 5

X will be a negative number. Not possible
Hence 5 = (7+x)/2
X = 3

Hence the third term is 3
Since the third term is 3, the second term would be 4 and the first term would be 2

Good approach. Understand it now. I got messed up after getting -ve number

Heman

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12 Aug 2006, 13:09
Got A i.e. 2 is the first term

_ _ _ 7 5

term 3 would be 5 = 7 + x or (7+x)/2 x = -2 or x = 3

hence x = 3 since list is of +ve integers

_ _ 3 7 5

7 = 3 + x i.e. x =4

_ 4 3 7 5

hence 3 = 4 + x or 3 = (4 + x)/2 which gives x = 2

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12 Aug 2006, 13:09
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# For a list of positive integers, from the third terms, each

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