Use Continuity and Momentum Relations to Find a2 and £2 in Terms of a1 and £1
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A sequence of terms a1, a2 ,a3, ..., a(m-1), am, is given by [#permalink] 
Updated on: 15 Dec 2020, 22:48
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A sequence of non-zero terms \(a_1\), \(a_2\), \(a_3\), ..., \(a_{m-1}\), \(a_m\), is given by \(a_k=(a_{k-1})^2(a_{k-2})\) for every k>2. If m=12, then how many terms in the given sequence are positive?
(1) \(a_3\) is positive
(2) \(a_4\) is positive
Originally posted by guerrero25 on 06 Apr 2013, 04:08.
Last edited by Bunuel on 15 Dec 2020, 22:48, edited 2 times in total.
Renamed the topic and edited the question.
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Re: A sequence of terms a1, a2 ,a3, ..., a(m-1), am, is given by [#permalink] 06 Apr 2013, 04:29
A sequence of non-zero terms \(a_1\), \(a_2\), \(a_3\), ..., \(a_{m-1}\), \(a_m\), is given by \(a_k=(a_{k-1})^2(a_{k-2})\) for every k>2. If m=12, then how many terms in the given sequence are positive?
From above:
\(a_3=(a_2)^2*a_1\);
\(a_4=(a_3)^2*a_2\);
...
(1) \(a_3\) is positive --> \(a_3=(a_2)^2*a_1=positive\) --> \(a_1=positive\). Now, if \(a_1=a_2=1\), then ALL 12 terms in the sequence will be positive but if \(a_1=1\), and \(a_2=-1\) (\(a_3=(a_2)^2*a_1=(-1)^2*1=1=positive\)), then not all the terms in the sequence will be positive. Not sufficient.
(2) \(a_4\) is positive --> \(a_4=(a_3)^2*a_2=positive\) --> \(a_2=positive\). The same here: if \(a_1=a_2=1\), then ALL 12 terms in the sequence will be positive but if \(a_1=-1\), and \(a_2=1\) (\(a_3=(a_2)^2*a_1=(1)^2*(-1)=-1\) and \(a_4=(a_3)^2*a_2=(-1)^2*1=1=positive\)), then not all the terms in the sequence will be positive. Not sufficient.
(1)+(2) From above we have that \(a_1=positive\) and \(a_2=positive\). Therefore, all 12 terms of the sequence are positive. Sufficient.
Answer: C.
Hope it's clear.
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Re: A sequence of terms a1, a2 ,a3, ..., a(m-1), am, is given by [#permalink] 07 Apr 2013, 09:40
Then.. The information for "every k>2" is irrelevant right
Posted from my mobile device
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Re: A sequence of terms a1, a2 ,a3, ..., a(m-1), am, is given by [#permalink] 07 Apr 2013, 21:33
marcovg4 wrote:
Then.. The information for "every k>2" is irrelevant right
Posted from my mobile device
"A sequence ... is given by \(a_k=(a_{k-1})^2(a_{k-2})\) for every k>2" means that the given formula applies for the terms starting \(a_3\).
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Re: A sequence of terms a1, a2 ,a3, ..., a(m-1), am, is given by [#permalink] 09 Apr 2013, 08:15
Bunuel wrote:
marcovg4 wrote:
Then.. The information for "every k>2" is irrelevant right
Posted from my mobile device
"A sequence ... is given by \(a_k=(a_{k-1})^2(a_{k-2})\) for every k>2" means that the given formula applies for the terms starting \(a_3\).
Oh I get it, thanks!
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Re: A sequence of terms a1, a2 ,a3, ..., a(m-1), am, is given by [#permalink] 06 Sep 2013, 11:12
Bunuel wrote:
A sequence of non-zero terms \(a_1\), \(a_2\), \(a_3\), ..., \(a_{m-1}\), \(a_m\), is given by \(a_k=(a_{k-1})^2(a_{k-2})\) for every k>2. If m=12, then how many terms in the given sequence are positive?
From above:
\(a_3=(a_2)^2*a_1\);
\(a_4=(a_3)^2*a_2\);
...
(1) \(a_3\) is positive --> \(a_3=(a_2)^2*a_1=positive\) --> \(a_1=positive\). Now, if \(a_1=a_2=1\), then ALL 12 terms in the sequence will be positive but if \(a_1=1\), and \(a_2=-1\) (\(a_3=(a_2)^2*a_1=(-1)^2*1=1=positive\)), then not all the terms in the sequence will be positive. Not sufficient.
(2) \(a_4\) is positive --> \(a_4=(a_3)^2*a_2=positive\) --> \(a_2=positive\). The same here: if \(a_1=a_2=1\), then ALL 12 terms in the sequence will be positive but if \(a_1=-1\), and \(a_2=1\) (\(a_3=(a_2)^2*a_1=(1)^2*(-1)=-1\) and \(a_4=(a_3)^2*a_2=(-1)^2*1=1=positive\)), then not all the terms in the sequence will be positive. Not sufficient.
(1)+(2) From above we have that \(a_1=positive\) and \(a_2=positive\). Therefore, all 12 terms of the sequence are positive. Sufficient.
Answer: C.
Hope it's clear.
Although the Answer is correct..but as I see the question Posted and the question in the image are different. Considering the question in the image a1 = +ve, a2=-ve, a3=+ve, a4=-ve and so on...Therefore, there will be 6 +ve terms in the sequence...
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Re: A sequence of terms a1, a2 ,a3, ..., a(m-1), am, is given by [#permalink] 18 Nov 2013, 10:51
Bunuel wrote:
Now, if \(a_1=a_2=1\), then ALL 12 terms in the sequence will be positive .
How can we say that because a1 and a2 are positive all terms will be positive?
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Re: A sequence of terms a1, a2 ,a3, ..., a(m-1), am, is given by [#permalink] 19 Nov 2013, 00:43
nikhil007 wrote:
Bunuel wrote:
Now, if \(a_1=a_2=1\), then ALL 12 terms in the sequence will be positive .
How can we say that because a1 and a2 are positive all terms will be positive?
\(a_3=(a_2)^2*a_1\);
\(a_4=(a_3)^2*a_2\);
...
Now, if a1 and a2 are both positive can a3 be negative? a4? an?
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Re: A sequence of terms a1, a2 ,a3, ..., a(m-1), am, is given by [#permalink] 19 Nov 2013, 22:58
guerrero25 wrote:
A sequence of non-zero terms \(a_1\), \(a_2\), \(a_3\), ..., \(a_{m-1}\), \(a_m\), is given by \(a_k=(a_{k-1})^2(a_{k-2})\) for every k>2. If m=12, then how many terms in the given sequence are positive?
(1) \(a_3\) is positive
(2) \(a_4\) is positive
My apologies . I could not find a way to type the sequence here , so I am attaching the DS question .
Statement 2 in the question and in the screenshot are different! is \(a_4\) positive or negative?
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Re: A sequence of terms a1, a2 ,a3, ..., a(m-1), am, is given by [#permalink] 20 Nov 2013, 00:32
emailmkarthik wrote:
guerrero25 wrote:
A sequence of non-zero terms \(a_1\), \(a_2\), \(a_3\), ..., \(a_{m-1}\), \(a_m\), is given by \(a_k=(a_{k-1})^2(a_{k-2})\) for every k>2. If m=12, then how many terms in the given sequence are positive?
(1) \(a_3\) is positive
(2) \(a_4\) is positive
My apologies . I could not find a way to type the sequence here , so I am attaching the DS question .
Statement 2 in the question and in the screenshot are different! is \(a_4\) positive or negative?
The discussion is on the question which says that \(a_4\) is positive.
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Re: A sequence of terms a1, a2 ,a3, ..., a(m-1), am, is given by [#permalink] 10 Sep 2015, 00:22
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Re: A sequence of terms a1, a2 ,a3, ..., a(m-1), am, is given by [#permalink]
10 Sep 2015, 00:22
Source: https://gmatclub.com/forum/a-sequence-of-terms-a1-a2-a3-a-m-1-am-is-given-by-150480.html
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