M14 Q3 - Why isn't 0 "a term"

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M14 Q3 - Why isn't 0 "a term" [#permalink]

15 Sep 2009, 00:48

A sequence is defined as follows:

An = n/n+1

How many of the first 100 terms of this sequence are less than 0.891?

A-7
B-8
C-9
D-10
E-12

I got this question wrong by considering and counting 0 as a term, yielding 0% as one of the terms below 0.891. I'm sure this is obvious to most, but how is one to know that 0 isnt a term when interpreting the question?

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Re: M14 Q3 - Why isn't 0 "a term" [#permalink]

15 Sep 2009, 17:34

dpgxxx wrote:

B. Go directly to 9th term which is 9/(9+1) = 0.9, which is slightly greater than 0.891. Smaller than 9th term is 8th term, which is 8/(8+1) = 0.889. So there are 8 terms in the sequence are less than 0.891.

Re: M14 Q3 - Why isn't 0 "a term"?

Thats a good question. Probably the counting starts from 1.
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Re: M14 Q3 - Why isn't 0 "a term" [#permalink]

05 Dec 2009, 14:28

Hm that is a terrible explanation. I did the same thing. Is this just a bad question?

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Re: M14 Q3 - Why isn't 0 "a term" [#permalink]

05 Dec 2009, 17:23

A sequence is a series of terms t1, t2, t3 up to tn. The sequence starts with t1, hence you start with n=1.

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Re: M14 Q3 - Why isn't 0 "a term" [#permalink]

05 Dec 2009, 21:13

gddunton wrote:

A sequence is a series of terms t1, t2, t3 up to tn. The sequence starts with t1, hence you start with n=1.

yes we can say a sequence is series of terms t1,t2 but that doesnt mean we start with n = 1. what if we take -ve integers to check?
lets consider this sequence S ={ -6,-4,-2,-1} here terms t1,t2 will correspond to -ve numbers

however found this discussion in the below link. Hope this helps

http://gmatclub.com/forum/m14-q3-why-isn-t-0-a-term-83934.html

Re: M14 Q3 - Why isn't 0 "a term" [#permalink] 05 Dec 2009, 21:13

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M14 Q3 - Why isn't 0 "a term"

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M14 Q3 - Why isn't 0 "a term"

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