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M15 #3

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Re: M15 #3 [#permalink]

07 Jan 2013, 06:20

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Marcab wrote:

Bunuel.
Since its given that n>1, then how can we even deal with A1.
This is why I got this one wrong.
Chose A, which is wrong.

It's not given that n>1. We are told that some sequence is defined such that A_{n+1}=\frac{A_{n}}{n+1} for all n>1.

So, this formula just gives values of the terms of the sequence starting from A_2.

Hope it's clear.
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Re: M15 #3 [#permalink]

07 Jan 2013, 06:27

Sorry for being dumb.
But then why we deal with even A1. For all n>1, the series must start with A2?
Regards.
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Re: M15 #3 [#permalink]

07 Jan 2013, 06:52

Marcab wrote:

Sorry for being dumb.
But then why we deal with even A1. For all n>1, the series must start with A2?
Regards.

The sequence starts with A1, naturally. But we have formula that gives the terms starting from A2. It's a common way of defining sequences, for example check these questions from MGMAT:
sequence-s-is-defined-as-sn-sn-1-1-1-sn-1-1-for-all-n-144382.html
sn-2sn-1-4-and-qn-4qn-1-8-for-all-n-1-if-s5-q4-125965.html

More questions on sequences to practice:
search.php?search_id=tag&tag_id=112 (PS)
search.php?search_id=tag&tag_id=111(DS)
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Re: M15 #3 [#permalink]

07 Jan 2013, 07:13

In fact I did such questions but it was not mentioned that n>1. Thats why there was no problem in doing questions where even finding other values was not at all confusing.
But I am unable to recover from the shock that this question gave.
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Re: M15 #3 [#permalink]

04 Feb 2013, 17:39

I was a bit confused on what the question is asking. I thought it was asking n could be a lot of number, and how many of those n could equal to 1/2...

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Re: M15 #3 [#permalink]

20 Feb 2013, 23:45

since, Nothing is given about "n" i.e. integer or whole no. or real no. etc.
so answer will depend upon nature of "n".

Re: M15 #3 [#permalink] 20 Feb 2013, 23:45

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M15 #3

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