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The nth term of the sequence a 1 , a 2 , a 3 , …, a n is def

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soumya1989
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GRE Math Challenge #10- The nth term of the sequence a 1 , a [#permalink] New post   05 Aug 2014, 03:49
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The nth term of the sequence \(a_{1}\) , \(a_{2}\) , \(a_{3}\) , …, an is defined as \(a_{n} = - (a_{n - 1} )\). The first term \(a_{1} = – 1\).

Quantity A
Quantity B
\(a_{5}\)
\(1\)


A. The quantity in Column A is greater
B. The quantity in Column B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given

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B
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13divyasha
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Re: GRE Math Challenge #10 [#permalink] New post   06 Aug 2014, 01:44
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Using a1 = -1, we get a2 = (-)a1 = 1
similarly a3 would be -1. Therefore, the odd nos. in the series will be (-)1 and even nos. will be 1.
Hence, a5 = (-)1. Column (B) is greater than Column (A).

The correct answer would be (B)
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wongpcla
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Re: GRE Math Challenge #10- The nth term of sequence [#permalink] New post   08 Dec 2017, 00:17
a odd are -1
a even are 1
so the answer is B
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The nth term of the sequence a 1 , a 2 , a 3 , …, a n is def [#permalink] New post   24 Apr 2018, 09:32
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This question is part of GREPrepClub - The Questions Vault Project



The nth term of the sequence \(a_{1}\) , \(a_{2}\) , \(a_{3}\) , …, an is defined as \(a_{n} = - (a_{n - 1} )\). The first term \(a_{1} = – 1\).

Quantity A
Quantity B
\(a_{5}\)
\(1\)


A. The quantity in Column A is greater
B. The quantity in Column B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given

Kudos for the right answer and solution.
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mohan514
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Re: The nth term of the sequence a 1 , a 2 , a 3 , …, a n is def [#permalink] New post   04 May 2018, 04:11
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On resolving we would get a sequence where all odd numbers in the series would be -1 and all even numbers would be +1. So a1, a3, a5, ..so on have values -1. a2, a4, a6, ... so on have +1.

Hence B is greater as a5 would have value of -1
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kruttikaaggarwal
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Re: The nth term of the sequence a 1 , a 2 , a 3 , …, a n is def [#permalink] New post   04 May 2018, 11:13
can someone please explain this. I do not understand how we solve the terms
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Re: The nth term of the sequence a 1 , a 2 , a 3 , …, a n is def [#permalink] New post   05 May 2018, 12:30
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It is given that \(a_{1}=-1\) and \(a_{n}=-a_{n-1}\).

So \(a_{2}=-a_{1}=1\).

And \(a_{3}=-a_{2}=-1\).

So we can see every odd term is -1 and every even term is +1. So \(a_{5]\) must be -1.

Hence option B is correct!
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adeel
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Re: The nth term of the sequence a 1 , a 2 , a 3 , …, a n is def [#permalink] New post   09 Jul 2018, 18:50
Carcass wrote:
The nth term of the sequence a 1 , a 2 , a 3 , …, an is defined as \(a_{n} = -(a_{n - 1} )\). The first term a1 equals –1.

Quantity A
Quantity B
a5
1


A. The quantity in Column A is greater
B. The quantity in Column B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given

Kudos for the right answer and solution.

My Answer is B
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Re: The nth term of the sequence a 1 , a 2 , a 3 , …, a n is def [#permalink] New post   10 Jul 2018, 08:22
The sequence: -1 1 -1 1 -1 -> a5 = -1 -> B.
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Hovsep
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Re: The nth term of the sequence a 1 , a 2 , a 3 , …, a n is def [#permalink] New post   25 Oct 2018, 01:37
thanks for the explanation
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Re: The nth term of the sequence a 1 , a 2 , a 3 , , a n is def [#permalink] New post   13 Aug 2022, 05:19
1
Basically

n_1 = -1
n_2 = (-1)^2
n_3 = (-1)^3

So whenever the term (=exponent) is odd = -1; whenever the term=exponent is even = 1.
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Re: The nth term of the sequence a 1 , a 2 , a 3 , , a n is def [#permalink]
13 Aug 2022, 05:19
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