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Intern  Joined: 22 Oct 2012
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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)

Originally posted by maglian on 19 Jan 2013, 11:53.
Last edited by Bunuel on 19 Jan 2013, 12:02, edited 1 time in total.
Edited the question and added OA.
Math Expert V
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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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Intern  Joined: 22 Oct 2012
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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.
Director  Joined: 29 Nov 2012
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Great Explantion! Thanks bunuel
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Re: A sequence of numbers (geometric sequence) is given by the  [#permalink]

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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
Intern  Joined: 20 May 2014
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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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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?
Manager  Joined: 28 Dec 2013
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BestGMATEliza wrote:
I have attached my notes on the problem.

Hope it helps!

Eliza Chute
Best GMAT Prep Courses

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

Sorry
Senior Manager  Joined: 08 Apr 2012
Posts: 320
Re: A sequence of numbers (geometric sequence) is given by the  [#permalink]

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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?
Math Expert V
Joined: 02 Sep 2009
Posts: 60647
Re: A sequence of numbers (geometric sequence) is given by the  [#permalink]

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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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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
Intern  Joined: 03 Jul 2015
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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
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Re: A sequence of numbers (geometric sequence) is given by the  [#permalink]

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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)

5 *(-1/2)^n - 5 *(-1/2)^n+1
((-1/2)^n - (-1/2)^n+1 ) < 1/2^3*5^3 (1/5)

one of will be -ve and other positive so we are looking at the summation.

1/2^n (1+1/2)

2^n = 1024*5*3/2 (16 = around 2*2*2)

n =13
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Why does |5∗(−1/2)^n− 5∗(−1/2)^n+1| becomes to/equals |5∗(−1/2)^n∗(1−(−1/2)^1)| ??

I don't understand why 5∗(−1/2)^n+1 can be expressed as n∗(1−(−1/2)^1

It would be nice if someone can explain. Veritas Prep GMAT Instructor V
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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)

Let's see what the sequence is like:

g(n)=5 *(-1/2)^n

g(1) = -5/2
g(2) = 5/4
g(3) = -5/8
g(4) = 5/16
etc

|g(n) - g(n+1)| would be the absolute value of difference between two such consecutive terms. Consecutive terms will be of opposite signs so they will actually get added. Since we are taking their absolute value, if they have a negative sign, it will become positive.

So all we need to think about is that the sum of absolute values of two consecutive terms should be less than 1/1000
1/1000 = 5/5000
2^12 = 4096

So the power of 2 should be more than 12.
5/2^13 + 5/2^14 = 5/8192 + 5/16384 = 15/16384
We see that if we divide 16384 by 15, we will get something more than 1000.

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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).

can you please clarify how to get $$2^n>7500$$?
Im stuck with $$\frac{15}{2}*(\frac{1}{2})^n<\frac{1}{1000}$$.

thank you!
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omavsp 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).

can you please clarify how to get $$2^n>7500$$?
Im stuck with $$\frac{15}{2}*(\frac{1}{2})^n<\frac{1}{1000}$$.

thank you!

$$\frac{15}{2}*(\frac{1}{2})^n<\frac{1}{1000}$$.

$$\frac{15*1000}{2}*(\frac{1}{2})^n<1$$.

$$7500*\frac{1}{2^n}<1$$.

$$7500<2^n$$.
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I tried this way.
|g(n)-g(n+1)|=1/1000=1*10^-3=0.0001;
g(1)=-5/2
g(2)=5/8
absolute difference between two values=|-5/2-5/8|=-15/8
similarly, g(10)=5/1024
g(11)=-5/2048
difference=|5/1024- (-5/2048)|=5/1024+5/2048=15/2048;

The numerator is always 15 for all 'n'
15/1000 will yield 3 decimal point. But to be of lower value than 0.0001, we need four zeroes after decimal.
To get that, 15 must be divided by a value greater than 15000;

possible value is 2^14. so Option is D)

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