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# 1) Series of A(n) is such that A(n) = A(n-1) / n. How many

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1) Series of A(n) is such that A(n) = A(n-1) / n. How many [#permalink]  18 Mar 2007, 18:26
1) Series of A(n) is such that A(n) = A(n-1) / n. How many elements of the series are bigger than 1/2?

(1) A(2) = 5
(2) A(1) - A(2) = 5

2) On the coordinate plane (6, 2) and (0, 6) are the endpoints of the diagonal of a square. What is the distance between point (0, 0) and the closest vertex of the square?

a) 1/sqrt (2) b) 1 c) sqrt (2) d) sqrt (3) e) 2*sqrt (3)
Senior Manager
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Q1. D
Stmt
Actually plugged in numbers
A2 = A1 / 2
so A1 = 10
A2 = 5
A3 = 5/3

Stmt 2
A1 - A2 = 5 implies A1 - (A1/2) = 5
Simplifying A1 = 10
Now we can again do actual calculations and answer the question asked.
Intern
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What about negative numbers? A(0), A(-1), etc.?
Senior Manager
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I clearly assumed n > 0.
I would say...n here needs to be greater than 0.
Because otherwise the whole series is undefined as a(0) is always undefined.
Which means question is ambiguous...

I would hate to see answer E here.
Whats OA?

I am working on your second questions...not quite hitting the right approach.
Intern
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The OA is D but the question did not mention that N > 0, hence my confusion.
Senior Manager
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Yes it is confusing.
Whats the source of this question? usually official questions are not ambiguous.
Director
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Either answer should be E or question shoud say only positive integers.

As from the calculation proposed by Kyatin

A(1) = A(1-1)/1 or A(1) = A(0) = 10...

also A(0) = A(-1)/0 or A(0) = infinity

A(0) cannot have two different values hence its insufficient.

regards,

Amardeep
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Well.... By convention, we consider a sequence with positive integers.... I would like to say that (D) is justified here ... Otherwise, we could even consider real numbers.... A(-sqrt(2)) = A(-sqrt(2)-1) / (-sqrt(2))... Makes not so much sense

a(-30) or a(-1) sounds like non sense, mathematically speaking ... Generally, we went to know from a sequence what happens if n tends to be infinite : convergent or divergent

Also, as the problem sets a(n) = a(n-1)/n, we cannot have n=0 because the equation simply exists and is definied ... But, we can have an admitted starting value A(0) for a sequence that serve to calculate the other terms.

All to say that we should consider n>0 and A(1) = A(0) a valid element
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Good insight fig,

Well its clearly written in GMAT quant section that all the numbers consider are real... though it can be negative integers as well, A(-1) which leads to A(0) = infinity doesnt make sense.

Also, we should understand that GMAT checks for aptitude i.e. avg maths which everybody should know... One shouldnt be Maths honors to solve these tricky but standard formula based probs.

regards,

Amardeep
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Amardeep Sharma wrote:
Good insight fig,

Well its clearly written in GMAT quant section that all the numbers consider are real... though it can be negative integers as well, A(-1) which leads to A(0) = infinity doesnt make sense.

Also, we should understand that GMAT checks for aptitude i.e. avg maths which everybody should know... One shouldnt be Maths honors to solve these tricky but standard formula based probs.

regards,

Amardeep

More about it here: http://en.wikipedia.org/wiki/Sequence.... I do agree that one could not know every mathematical term used but ok... Once one of these terms is used, it conveys a certain meaining that we cannot avoid ... Here, "serie" implies n >= 0

Hope that helps

I speak in terms of pure mathematical logic here

Also, notice that the GMAT would give the exact domain of definition for n ... That is, in a sens, missing in the quesiton to be a "whole" GMAT one
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What is the answer to the following question?

**********************
1/(2 â€“ sqrt(3)) = ?

(a) sqrt(3) - 2
(b) 2 + sqrt(3)
(c) sqrt(2) + sqrt(3)
(d) 2 - sqrt(3)
(e) sqrt(3) + 4
**********************
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1/(2 â€“ sqrt(3)) = ?

We have rationalize the denominator. Multiply both numerator and denominator by (2 + sqrt(3))

1/(2 â€“ sqrt(3))

= (2 + sqrt(3))/[(2 + sqrt(3))*(2 â€“ sqrt(3))]

The denominator is now of the form (a+b)*(a-b) = a^2-b^2

Hence

1/(2 â€“ sqrt(3))

= (2 + sqrt(3))/[(2 + sqrt(3))*(2 â€“ sqrt(3))]

= (2 + sqrt(3))/[2^2 -(sqrt(3))^2]

= (2 + sqrt(3))/[4-3]

= (2 + sqrt(3))

Hence B
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...In the term a(n)...n represents the number of the term ..a sequence will have finite/infinite number of terms ..so when we are talking about the first term of the sequence ...n = 1....hence n cannot be zero ..(since always first term means n=1)....
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Answer to the second question of the post http://www.gmatclub.com/phpbb/viewtopic.php?t=43543
Senior Manager
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kyatin wrote:
Q1. D
Stmt
Actually plugged in numbers
A2 = A1 / 2
so A1 = 10
A2 = 5
A3 = 5/3

Stmt 2
A1 - A2 = 5 implies A1 - (A1/2) = 5
Simplifying A1 = 10
Now we can again do actual calculations and answer the question asked.

Kyatin, could you please explain it again? I am sorry I could not get it.

Thanks.
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Summer3

Which step did you not get?

Let me know and I can elaborate.
Senior Manager
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all of it
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The answer to the second question - the distance between the origin and the nearest vertex of square is sqrt(2).

Solving the equations lead to the nearest vertex as (1,1).
Distance between (0,0) and (1,1) is sqrt(2).

Approach,
Let the nearest vertex is (x,y)
the length of the diagonal is = distance between (0,6) and (6,2) = sqrt(52). Hence the length of each side of the square = sqrt(26)
Now the distance between the each of the known vertices and the nearest vertex should be the length of a side. Equalise them will end up in the equation: 3x-2y =1
distance between (0,6) and (x,y) is sqrt(26), solving this y = 1 or 7. Putting the values of y in 3x-2y =1 , possible values of x are 1, 5.
Hence the nearest vertex is (1,1).
distance between (0,0) and (1,1) is sqrt(2).

Can somebody suggests me the quicker approach.
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vshaunak@gmail.com wrote:
The answer to the second question - the distance between the origin and the nearest vertex of square is sqrt(2).

Solving the equations lead to the nearest vertex as (1,1).
Distance between (0,0) and (1,1) is sqrt(2).

Approach,
Let the nearest vertex is (x,y)
the length of the diagonal is = distance between (0,6) and (6,2) = sqrt(52). Hence the length of each side of the square = sqrt(26)
Now the distance between the each of the known vertices and the nearest vertex should be the length of a side. Equalise them will end up in the equation: 3x-2y =1
distance between (0,6) and (x,y) is sqrt(26), solving this y = 1 or 7. Putting the values of y in 3x-2y =1 , possible values of x are 1, 5.
Hence the nearest vertex is (1,1).
distance between (0,0) and (1,1) is sqrt(2).

Can somebody suggests me the quicker approach.

If u are fine with vectors ... Have a look at my post : http://www.gmatclub.com/phpbb/viewtopic.php?t=43543
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