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
19 Jan 2013, 11:53
6
This post received KUDOS
28
This post was BOOKMARKED
Question Stats:
34% (04:48) correct
66% (02:35) wrong based on 942 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.
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: 39697

Re: A sequence of numbers (geometric sequence) is given by the [#permalink]
Show Tags
19 Jan 2013, 12:17
11
This post received KUDOS
Expert's post
12
This post was BOOKMARKED
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.
_________________
New to the Math Forum? Please read this: All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



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: 878

Re: A sequence of numbers (geometric sequence) is given by the [#permalink]
Show Tags
20 May 2013, 07:04
Great Explantion! Thanks bunuel
_________________
Click +1 Kudos if my post helped...
Amazing Free video explanation for all Quant questions from OG 13 and much more http://www.gmatquantum.com/og13th/
GMAT Prep software What if scenarios http://gmatclub.com/forum/gmatprepsoftwareanalysisandwhatifscenarios146146.html



Manager
Joined: 07 Apr 2012
Posts: 126
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: 4

Re: A sequence of numbers (geometric sequence) is given by the [#permalink]
Show Tags
22 May 2014, 11:45
1
This post received KUDOS
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: 39

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: 80

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: 453

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: 39697

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
_________________
New to the Math Forum? Please read this: All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Manager
Joined: 10 Jun 2015
Posts: 128

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: 34

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: 376
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




Re: A sequence of numbers (geometric sequence) is given by the
[#permalink]
29 Aug 2016, 02:08







