Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 22 Oct 2012
Posts: 16

A sequence of numbers (geometric sequence) is given by the
[#permalink]
Show Tags
Updated on: 19 Jan 2013, 12:02
Question Stats:
31% (03:11) correct 69% (03:01) wrong based on 809 sessions
HideShow timer Statistics
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)
Official Answer and Stats are available only to registered users. Register/ Login.
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
Joined: 02 Sep 2009
Posts: 60647

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




Intern
Joined: 22 Oct 2012
Posts: 16

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



Director
Joined: 29 Nov 2012
Posts: 682

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



Manager
Joined: 07 Apr 2012
Posts: 87
Location: United States
Concentration: Entrepreneurship, Operations
GPA: 3.9
WE: Operations (Manufacturing)

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



Intern
Joined: 20 May 2014
Posts: 3

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



Intern
Joined: 20 May 2014
Posts: 31

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



Manager
Joined: 28 Dec 2013
Posts: 65

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



Senior Manager
Joined: 08 Apr 2012
Posts: 320

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



Math Expert
Joined: 02 Sep 2009
Posts: 60647

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



Manager
Joined: 10 Jun 2015
Posts: 110

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



Intern
Joined: 03 Jul 2015
Posts: 27

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



Senior Manager
Joined: 07 Sep 2014
Posts: 336
Concentration: Finance, Marketing

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



Intern
Joined: 22 Oct 2017
Posts: 30

A sequence of numbers (geometric sequence) is given by the
[#permalink]
Show Tags
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.



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 10011
Location: Pune, India

A sequence of numbers (geometric sequence) is given by the
[#permalink]
Show Tags
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)
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Intern
Joined: 20 Aug 2017
Posts: 20

Re: A sequence of numbers (geometric sequence) is given by the
[#permalink]
Show Tags
30 Jun 2019, 05:53
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. 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!



Math Expert
Joined: 02 Sep 2009
Posts: 60647

Re: A sequence of numbers (geometric sequence) is given by the
[#permalink]
Show Tags
30 Jun 2019, 05:57
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) \(g(n)  g(n+1) =5 *(\frac{1}{2})^n5 *(\frac{1}{2})^{n+1}\) > 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. 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! \(\frac{15}{2}*(\frac{1}{2})^n<\frac{1}{1000}\). \(\frac{15*1000}{2}*(\frac{1}{2})^n<1\). \(7500*\frac{1}{2^n}<1\). \(7500<2^n\).
_________________



Intern
Joined: 12 Feb 2018
Posts: 18

Re: A sequence of numbers (geometric sequence) is given by the
[#permalink]
Show Tags
06 Sep 2019, 03:32
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=5/25/8=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) Please correct me if am wrong




Re: A sequence of numbers (geometric sequence) is given by the
[#permalink]
06 Sep 2019, 03:32






