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A sequence of numbers (geometric sequence) is given by the
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Updated on: 19 Jan 2013, 12:02
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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)
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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.




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Re: A sequence of numbers (geometric sequence) is given by the
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19 Jan 2013, 12:17
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) \(g(n)  g(n+1) =5 *(\frac{1}{2})^n5 *(\frac{1}{2})^{n+1}\) > factor out \(5 *(\frac{1}{2})^n\): \(5 *(\frac{1}{2})^n5 *(\frac{1}{2})^{n+1}=5 *(\frac{1}{2})^n*(1(\frac{1}{2})^1)=5 *(\frac{1}{2})^n*\frac{3}{2}=\frac{15}{2}*(\frac{1}{2})^n\). 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
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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).
Answer: D.
thank you. very simple explanation. you rock.



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Re: A sequence of numbers (geometric sequence) is given by the
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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
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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
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22 May 2014, 11:45
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
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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) \(g(n)  g(n+1) =5 *(\frac{1}{2})^n5 *(\frac{1}{2})^{n+1}\) > factor out \(5 *(\frac{1}{2})^n\): \(5 *(\frac{1}{2})^n5 *(\frac{1}{2})^{n+1}=5 *(\frac{1}{2})^n*(1(\frac{1}{2})^1)=5 *(\frac{1}{2})^n*\frac{3}{2}=\frac{15}{2}*(\frac{1}{2})^n\). 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
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02 Jul 2014, 21:24
BestGMATEliza wrote: I have attached my notes on the problem. Hope it helps! Eliza Chute Best GMAT Prep CoursesYour one stop for all your GMAT studying needs! 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
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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) \(g(n)  g(n+1) =5 *(\frac{1}{2})^n5 *(\frac{1}{2})^{n+1}\) > factor out \(5 *(\frac{1}{2})^n\): \(5 *(\frac{1}{2})^n5 *(\frac{1}{2})^{n+1}=5 *(\frac{1}{2})^n*(1(\frac{1}{2})^1)=5 *(\frac{1}{2})^n*\frac{3}{2}=\frac{15}{2}*(\frac{1}{2})^n\). 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. 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
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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) \(g(n)  g(n+1) =5 *(\frac{1}{2})^n5 *(\frac{1}{2})^{n+1}\) > factor out \(5 *(\frac{1}{2})^n\): \(5 *(\frac{1}{2})^n5 *(\frac{1}{2})^{n+1}=5 *(\frac{1}{2})^n*(1(\frac{1}{2})^1)=5 *(\frac{1}{2})^n*\frac{3}{2}=\frac{15}{2}*(\frac{1}{2})^n\). 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. 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 \(\frac{15}{2}*(\frac{1}{2})^n\). Also, PLEASE read this: rulesforpostingpleasereadthisbeforeposting133935.html#p1096628
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Re: A sequence of numbers (geometric sequence) is given by the
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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 so the answer is (D)



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



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Re: A sequence of numbers (geometric sequence) is given by the
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29 Aug 2016, 02:08
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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A sequence of numbers (geometric sequence) is given by the
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08 Nov 2017, 05:10
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.



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A sequence of numbers (geometric sequence) is given by the
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08 Nov 2017, 05:28
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. Answer (D)
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