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Old 2004 December 30th, 02:54 AM   #1 (permalink)
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Exclamation Hard one

In the infinite sequence A, ,
where x is a positive integer constant. For what value of n is the ratio
of to equal to ?

(A) 8
(B) 7
(C) 6
(D) 5
(E) 4

source: Manhattan GMAT
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Old 2004 December 30th, 04:22 AM   #2 (permalink)
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Re: Hard one

is it B ? n = 7
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Old 2005 January 5th, 01:39 PM   #3 (permalink)
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Re: Hard one

I most humbly and respectfully would go for B too... whats the answer..!
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Old 2005 January 7th, 12:15 PM   #4 (permalink)
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Re: Hard one

I get answer of 7, or B, too. How would you approach this problem in general other than plugging in the answers after expanding the denominator?
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Old 2005 January 7th, 08:20 PM   #5 (permalink)
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Re: Hard one

any more thoughts on this one
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Old 2005 January 7th, 10:40 PM   #6 (permalink)
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Re: Hard one

The method I followed was to reduce the Q to (x^6/ x ) * ( Y/Y)
the eqn An= (x ^ n -1)(1+ x + x^2 + x^3 +.... ) ----------------------------(1)

and the eqn x(1+x(1+x(1+x...))))
which i call Z can be reduced to x( 1+x+x^2+x^3 ..) --------(2)

from (1) and (2) we get An / Z = x^(n-1) / x
therefore for getting answer x^5 (n-1) = 6
therefore n=7

answer: B
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Old 2005 February 4th, 07:45 PM   #7 (permalink)
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Re: Hard one

Its B.
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Old 2005 February 5th, 05:30 AM   #8 (permalink)
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Re: Hard one

b for me too
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Old 2005 February 11th, 03:29 AM   #9 (permalink)
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Re: Hard one

Quote:
Originally Posted by Sargataur
any more thoughts on this one
Sargataur, is this the type of question one would encounter on the GMAT?
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Old 2005 March 21st, 10:45 AM   #10 (permalink)
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Re: Hard one

[quote=Stormgal]In the infinite sequence A, ,
where x is a positive integer constant. For what value of n is the ratio
of to equal to ?

(A) 8
(B) 7
(C) 6
(D) 5
(E) 4

the answer is B
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