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A sequence of numbers (geometric sequence) is given by the

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A sequence of numbers (geometric sequence) is given by the [#permalink] New post 19 Jan 2013, 10:53
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A sequence of numbers (geometric sequence) is given by the expression: g(n)=5 *(-1/2)^n. If the sequence begins with n = 1, what are the first two terms for which |g(n) - g(n+1)| < 1/1000 ?

A. g(10), g(11)
B. g(11), g(12)
C. g(12), g(13)
D. g(13), g(14)
E. g(14), g(15)
[Reveal] Spoiler: OA

Last edited by Bunuel on 19 Jan 2013, 11:02, edited 1 time in total.
Edited the question and added OA.
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Re: A sequence of numbers (geometric sequence) is given by the [#permalink] New post 19 Jan 2013, 11:17
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maglian wrote:
A sequence of numbers (geometric sequence) is given by the expression: g(n)=5 *(-1/2)^n. If the sequence begins with n = 1, what are the first two terms for which |g(n) - g(n+1)| < 1/1000 ?

A. g(10), g(11)
B. g(11), g(12)
C. g(12), g(13)
D. g(13), g(14)
E. g(14), g(15)


|g(n) - g(n+1)| =|5 *(-\frac{1}{2})^n-5 *(-\frac{1}{2})^{n+1}| --> factor out 5 *(-\frac{1}{2})^n:

|5 *(-\frac{1}{2})^n-5 *(-\frac{1}{2})^{n+1}|=|5 *(-\frac{1}{2})^n*(1-(-\frac{1}{2})^1)|=|5 *(-\frac{1}{2})^n*\frac{3}{2}|=\frac{15}{2}*(\frac{1}{2})^n.

So, we need to find the least value of n for which \frac{15}{2}*(\frac{1}{2})^n<\frac{1}{1000} --> 2^n>7500 --> the least n for which this inequality hods true is 13 (2^13=~8000>7500).

Thus, the first two terms, g(n) and g(n+1), for which the given inequality holds true are g(13) and g(14).

Answer: D.
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Re: A sequence of numbers (geometric sequence) is given by the [#permalink] New post 19 Jan 2013, 11:23
Bunuel wrote:

So, we need to find the least value of n for which \frac{15}{2}*(\frac{1}{2})^n<\frac{1}{1000} --> 2^n>7500 --> the least n for which this inequality hods true is 13 (2^13=~8000>7500).

Thus, the first two terms, g(n) and g(n+1), for which the given inequality holds true are g(13) and g(14).

Answer: D.


thank you. very simple explanation. you rock.
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Re: A sequence of numbers (geometric sequence) is given by the [#permalink] New post 20 May 2013, 06:04
Great Explantion! Thanks bunuel
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Re: A sequence of numbers (geometric sequence) is given by the [#permalink] New post 11 Oct 2013, 12:06
Though I got it wrong due to wrong approximation, I found an alternate method.

Take n=1, so G(1) - G(2) = 5(-1/2)^1 - 5(-1/2)^2.
On solving we get some number upon power of 2 as 2( since 2^2 is LCM)

Since the value compared is with 1000, we need to take values of 2 above index 10

Take option C:
| G(12) - G(13) | = 5(-1/2)^12 - 5(-1/2)^13 , solve it to get 5/2^13. If we take split denominator as 2^3 and 2^10
we get (5/2^3)*(1/2^10).

5/8 = 0.625 and 1/2^10 = 1024 , if you compare it with RHS 1/1000 we are almost there. I had approximated 1024 as 1000, and wrongly took 5/8 as 0.125 which gave me it less than 1000, but in actuals its more than 1/1000
So next number could be the answer, check for G13 and G14, you have the answer.
Hope I am able to express it clearly

Honestly speaking I dont think such questions can come up in GMAT
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Re: A sequence of numbers (geometric sequence) is given by the [#permalink] New post 22 May 2014, 10:45
We can write g(n+1) = -1/2*g(n)

so g(n) - g(n+1) = g(n) +1/2*g(n) = 3/2 g(n)

3/2 g(n) < 1/1000

or g(n)<1/1500

(1/2)^n<1/7500 (ignoring the -ve sign due to modulus)

n=13 is the least value which satisfies the equation.

Thanks,
Chirag Bhagat
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Re: A sequence of numbers (geometric sequence) is given by the [#permalink] New post 02 Jul 2014, 19:01
Bunuel wrote:
maglian wrote:
A sequence of numbers (geometric sequence) is given by the expression: g(n)=5 *(-1/2)^n. If the sequence begins with n = 1, what are the first two terms for which |g(n) - g(n+1)| < 1/1000 ?

A. g(10), g(11)
B. g(11), g(12)
C. g(12), g(13)
D. g(13), g(14)
E. g(14), g(15)


|g(n) - g(n+1)| =|5 *(-\frac{1}{2})^n-5 *(-\frac{1}{2})^{n+1}| --> factor out 5 *(-\frac{1}{2})^n:

|5 *(-\frac{1}{2})^n-5 *(-\frac{1}{2})^{n+1}|=|5 *(-\frac{1}{2})^n*(1-(-\frac{1}{2})^1)|=|5 *(-\frac{1}{2})^n*\frac{3}{2}|=\frac{15}{2}*(\frac{1}{2})^n.

So, we need to find the least value of n for which \frac{15}{2}*(\frac{1}{2})^n<\frac{1}{1000} --> 2^n>7500 --> the least n for which this inequality hods true is 13 (2^13=~8000>7500).

Thus, the first two terms, g(n) and g(n+1), for which the given inequality holds true are g(13) and g(14).

Answer: D.



HOW DID YOU GET TO 2^(n) < 7500 exactly?
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Re: A sequence of numbers (geometric sequence) is given by the [#permalink] New post 02 Jul 2014, 20:24
BestGMATEliza wrote:
I have attached my notes on the problem.

Hope it helps!

Eliza Chute
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This was great but where does 2 * 2^n come from?

Sorry
Re: A sequence of numbers (geometric sequence) is given by the   [#permalink] 02 Jul 2014, 20:24
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