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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]

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19 Jan 2013, 11:53
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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, 12:02, 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]

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19 Jan 2013, 12:17
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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).

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Re: A sequence of numbers (geometric sequence) is given by the [#permalink]

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19 Jan 2013, 12:23
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).

thank you. very simple explanation. you rock.
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Re: A sequence of numbers (geometric sequence) is given by the [#permalink]

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20 May 2013, 07:04
Great Explantion! Thanks bunuel
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Re: A sequence of numbers (geometric sequence) is given by the [#permalink]

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11 Oct 2013, 13:06
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]

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22 May 2014, 11:45
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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]

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02 Jul 2014, 20:01
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).

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]

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02 Jul 2014, 21:24
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
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Re: A sequence of numbers (geometric sequence) is given by the [#permalink]

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31 Aug 2014, 07:50
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).

Hi Bunuel,
I got answer C, and also backed it up by calculation with a calculator.
it comes down to |-5*(1/2)^n +5*(1/2)*(n+1)| = |-5*(1/2)^(n+1)| < 1/1000
if (n+1) = 13, we get the answer we were looking for, so the answer should be C.
Are you sure that the answer needs to be D?
Can you find something wrong with my logic/calculations?
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Re: A sequence of numbers (geometric sequence) is given by the [#permalink]

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01 Sep 2014, 05:43
ronr34 wrote:
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).

Hi Bunuel,
I got answer C, and also backed it up by calculation with a calculator.
it comes down to |-5*(1/2)^n +5*(1/2)*(n+1)| = |-5*(1/2)^(n+1)| < 1/1000
if (n+1) = 13, we get the answer we were looking for, so the answer should be C.
Are you sure that the answer needs to be D?
Can you find something wrong with my logic/calculations?

You cannot factor out -1 the way you did. (-1/2)^n does not equal to -1*(1/2)^n if n is even, and we don't know whether it's even or odd.

The whole expression comes down to $$\frac{15}{2}*(\frac{1}{2})^n$$.

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Re: A sequence of numbers (geometric sequence) is given by the [#permalink]

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14 Jun 2015, 05:39
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)

the sequence is 15/4, 15/8, 15/16,......,15/2048, 15/4096, 15/ 8192, 15/16384 and so on
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Re: A sequence of numbers (geometric sequence) is given by the [#permalink]

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15 Feb 2016, 00:42
In the last step how did you come to the figure 2^13?
how do we arrive at 2 raised to the power of such a big number? esp the choice gives us close options
Re: A sequence of numbers (geometric sequence) is given by the   [#permalink] 15 Feb 2016, 00:42
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