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by jimmiejaz » Wed Dec 03, 2008 2:24 pm
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?

7
8
9
10
12

OA B
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Re: Good question

by Stuart@KaplanGMAT » Wed Dec 03, 2008 2:35 pm
jimmiejaz wrote: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?

7
8
9
10
12
Let's jot down the first few terms to understand the sequence.

if n=1, we get 1/2
if n=2, we get 2/3
if n=3, we get 3/4.

So we need to find the first fraction in the sequence that's greater than .891.

Let's start with choice (a).

7/8 = .875... still under
8/9 = .888.. not there yet!
9/10 = .9. Yay, we're past our mark. Since our fractions are getting larger and larger, we now know that only the first 8 terms are less than .891. Choose (B).
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by jimmiejaz » Wed Dec 03, 2008 2:37 pm
Thanks for the explantion.
I got this question because i started with A0.
How will be sure that this sequence doesnt start from 0?
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by Stuart@KaplanGMAT » Wed Dec 03, 2008 2:49 pm
jimmiejaz wrote:Thanks for the explantion.
I got this question because i started with A0.
How will be sure that this sequence doesnt start from 0?
Sets start with the 1st term, not the 0th term (which doesn't make any sense :D ).
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by jimmiejaz » Wed Dec 03, 2008 2:57 pm
Stuart Kovinsky wrote:
jimmiejaz wrote:Thanks for the explantion.
I got this question because i started with A0.
How will be sure that this sequence doesnt start from 0?
Sets start with the 1st term, not the 0th term (which doesn't make any sense :D ).
Stuart , am still not able to follow this. When we study A.P or a G.P we start with A0. If, A0=1, An=n/2(2A0+(n-1)d).
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by ifairo » Wed Dec 03, 2008 3:42 pm
Easy way is to start from the value 0.89. This means 9/10 approx right. n = 9, and n+1 = 10. So, the first 8 terms.
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by cramya » Wed Dec 03, 2008 4:19 pm
Stuart , am still not able to follow this. When we study A.P or a G.P we start with A0. If, A0=1, An=n/2(2A0+(n-1)d).
I could be mistaken here but I have always seen in an AP

Sn = n/2(2*a1 + (n-1)d)

a1->first term

If it was 0

then an=a+(n-1)d will be a0 = a-d which will not make sense IMO.
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by jimmiejaz » Wed Dec 03, 2008 4:28 pm
my bad!!!!!!
U r absolutely true.
A0 cant be A0-d the eqn will not hold.
I got confused big time.....
Thanks for making it clear.
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by logitech » Wed Dec 03, 2008 5:52 pm
N/N+1 > 891/1000

1000N>891N+891

108N>891

N>8,blablabla

So N(max) = 8

Choose (B)
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