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# In a sequence of positive integers

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BSchool Forum Moderator
Joined: 12 Aug 2015
Posts: 2642
GRE 1: 323 Q169 V154
In a sequence of positive integers [#permalink]

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23 Apr 2017, 07:50
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Difficulty:

75% (hard)

Question Stats:

53% (01:11) correct 47% (01:49) wrong based on 83 sessions

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In a sequence of positive integers, $$A(n)$$,the the terms is defined as $$[A(n-1) - 1]^2$$. If 9 is one of the terms of the sequence,then what are the two terms immediately next to the 9?

A){64,63^2}
B){9,63}
C){4,64}
D){4,9}
E){63,64}

Source =>NOVA.

P.S=> This might turn out to be a poor quality question.I feel it has language issues

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BSchool Forum Moderator
Joined: 12 Aug 2015
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GRE 1: 323 Q169 V154
Re: In a sequence of positive integers [#permalink]

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23 Apr 2017, 07:51
1
Hi Bunuel
Here is my doubt =>
In the OE the writer is assuming that the word immediately next means the closest vicinity.
Hence he calculated the previous term and the next term instead of the next two terms.
I don't think he is correct with that mindset. Shouldn't the term immediately next mean the next two terms?

Regards
Stone Cold

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Re: In a sequence of positive integers [#permalink]

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23 Apr 2017, 18:48
1
I think this is a poor quality question. It shouldn't show any confusion within the question.
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Re: In a sequence of positive integers [#permalink]

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20 May 2017, 03:43
1
Ideally the question should ask: 'What are the two terms immediately adjacent to 9'.

Here we know that a2 = (a1-1)^2, a3 = (a2-1)^2.. and so on..

so the term immediately after 9 = (9-1)^2 = 8^2 = 64

Let the term previous to 9 be x.
9 = (x-1)^2 or x-1 = 3 Or x-1= -3
which gives two values of x: x = 4 or x = -2
But since its given that its a sequence of positive integers, we will ignore -2. Thus the previous term is 4.

So the two terms immediately adjacent to 9 are : 4 and 64

Re: In a sequence of positive integers   [#permalink] 20 May 2017, 03:43
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# In a sequence of positive integers

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