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# A sequence of terms a1, a2 ,a3, ..., a(m-1), am, is given by

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Re: A sequence of terms a1, a2 ,a3, ..., a(m-1), am, is given by [#permalink]
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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]
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]
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.

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]
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]
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]
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]
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]
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Re: A sequence of terms a1, a2 ,a3, ..., a(m-1), am, is given by [#permalink]
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