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M15 #3

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Re: M15 #3 [#permalink] New post 07 Jan 2013, 05:20
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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] New post 07 Jan 2013, 05:27
Expert's post
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] New post 07 Jan 2013, 05:52
Expert's post
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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PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: M15 #3 [#permalink] New post 07 Jan 2013, 06:13
Expert's post
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] New post 04 Feb 2013, 16:39
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] New post 20 Feb 2013, 22:45
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since, Nothing is given about "n" i.e. integer or whole no. or real no. etc.
so answer will depend upon nature of "n".
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Re: M15 #3 [#permalink] New post 20 Apr 2014, 07:21
Elements of this series gradually get smaller when n increases. Knowing 1 element would let us calculate the others, so definitely we will know how many is smaller than 1/2
1) is sufficient. 2) lets us to know the value of A2, so also sufficient.

Choose d
Re: M15 #3   [#permalink] 20 Apr 2014, 07:21
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