Find all School-related info fast with the new School-Specific MBA Forum

It is currently 30 Oct 2014, 21:39

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

For every integer m from 1 to 100, inclusive, the mth term

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
1 KUDOS received
Director
Director
avatar
Status: Preparing for the 4th time -:(
Joined: 25 Jun 2011
Posts: 560
Location: United Kingdom
Concentration: International Business, Strategy
GMAT Date: 06-22-2012
GPA: 2.9
WE: Information Technology (Consulting)
Followers: 13

Kudos [?]: 562 [1] , given: 217

For every integer m from 1 to 100, inclusive, the mth term [#permalink] New post 04 Mar 2012, 15:18
1
This post received
KUDOS
3
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

51% (03:06) correct 49% (01:24) wrong based on 121 sessions
For every integer m from 1 to 100, inclusive, the mth term of a certain sequence is given by (-1)^m*2^(-m). If N is the sum of the fi…rst 100 terms in the sequence, then N is

(A) less than -1
(B) between -1 and -1/2
(C) between -1/2 and 0
(D) between 0 and 1/2
(E) greater than 1/2

Any idea how solve this question? I have got no clue how to.
[Reveal] Spoiler: OA

_________________

Best Regards,
E.

MGMAT 1 --> 530
MGMAT 2--> 640
MGMAT 3 ---> 610 :-(

Expert Post
3 KUDOS received
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 23499
Followers: 3636

Kudos [?]: 29392 [3] , given: 2924

Re: For every integer m from 1 to 100, inclusive, the mth term [#permalink] New post 04 Mar 2012, 16:24
3
This post received
KUDOS
Expert's post
1
This post was
BOOKMARKED
enigma123 wrote:
For every integer m from 1 to 100, inclusive, the mth term of a certain sequence is given by (-1)^m*2^(-m). If N is the sum of the fi…rst 100 terms in the sequence, then N is

(A) less than -1
(B) between -1 and -1/2
(C) between -1/2 and 0
(D) between 0 and 1/2
(E) greater than 1/2

Any idea how solve this question? I have got no clue how to.


1st term (-1)^1*2^{-1}=-\frac{1}{2};
2nd term (-1)^2*2^{-2}=\frac{1}{4};
3rd term (-1)^3*2^{-3}=-\frac{1}{8};
4th term (-1)^4*2^{-4}=\frac{1}{16};
5th term (-1)^5*2^{-5}=-\frac{1}{32};
6th term (-1)^6*2^{-6}=\frac{1}{64};
...

The sum of 1st and 2nd = -\frac{1}{4};
The sum of 3rd and 4th = -\frac{1}{16};
The sum of 5th and 6th = -\frac{1}{64};
...

So, we should basically sum 100/2=50 terms which form a geometric progression with the first term of -\frac{1}{4} and common ratio of \frac{1}{4}: -\frac{1}{4}, -\frac{1}{16}, -\frac{1}{64}, ...

The sum of infinite geometric progression with common ratio |r|<1, is sum=\frac{b}{1-r}, where b is the first term.

So, in our case, since the # of terms is large enough, the sum will be very close to sum=\frac{-\frac{1}{4}}{1-\frac{1}{4}}=-\frac{1}{3}.

Answer: C.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

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?
25 extra-hard Quant Tests

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Senior Manager
Senior Manager
avatar
Joined: 16 Feb 2012
Posts: 260
Concentration: Finance, Economics
Followers: 4

Kudos [?]: 79 [0], given: 121

GMAT ToolKit User
Re: For every integer m from 1 to 100, inclusive, the mth term [#permalink] New post 05 May 2012, 04:40
I don't get it. Isn't the first term -1/2 and the common ratio 1/2?
And why do you need to sum up 1 & 2, 3 & 4, 5 & 6 term... etc.?
_________________

Kudos if you like the post!

Failing to plan is planning to fail.

Intern
Intern
User avatar
Joined: 30 Mar 2012
Posts: 36
Followers: 0

Kudos [?]: 2 [0], given: 11

Re: For every integer m from 1 to 100, inclusive, the mth term [#permalink] New post 19 Jun 2012, 04:09
Stiv wrote:
I don't get it. Isn't the first term -1/2 and the common ratio 1/2?
And why do you need to sum up 1 & 2, 3 & 4, 5 & 6 term... etc.?


Well both the solution would fetch you the same result of -1/3 if you consider it as an infinite series..
and that would solve the question.
_________________

This time its personal..

Expert Post
1 KUDOS received
Veritas Prep GMAT Instructor
User avatar
Joined: 16 Oct 2010
Posts: 4887
Location: Pune, India
Followers: 1161

Kudos [?]: 5435 [1] , given: 165

Re: For every integer m from 1 to 100, inclusive, the mth term [#permalink] New post 19 Jun 2012, 05:00
1
This post received
KUDOS
Expert's post
enigma123 wrote:
For every integer m from 1 to 100, inclusive, the mth term of a certain sequence is given by (-1)^m*2^(-m). If N is the sum of the fi…rst 100 terms in the sequence, then N is

(A) less than -1
(B) between -1 and -1/2
(C) between -1/2 and 0
(D) between 0 and 1/2
(E) greater than 1/2

Any idea how solve this question? I have got no clue how to.


I have discussed this question in details using both the methods - GP perspective and non GP perspective.
Check it out here:
http://www.veritasprep.com/blog/2012/04 ... rspective/
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Save $100 on Veritas Prep GMAT Courses And Admissions Consulting
Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options.

Veritas Prep Reviews

Intern
Intern
avatar
Joined: 11 Jul 2012
Posts: 46
Followers: 0

Kudos [?]: 0 [0], given: 0

Re: For every integer m from 1 to 100, inclusive, the mth term [#permalink] New post 30 Oct 2012, 22:37
Why should we consider this as an infinite sequence?
Expert Post
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 23499
Followers: 3636

Kudos [?]: 29392 [0], given: 2924

Re: For every integer m from 1 to 100, inclusive, the mth term [#permalink] New post 30 Oct 2012, 22:43
Expert's post
Ousmane wrote:
Why should we consider this as an infinite sequence?


We are not saying that it's an infinite sequence. But if it were, then the sum would be -1/3 and since we have the sum of the large enough number of terms (100), then the actual sum would be very close to -1/3.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

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?
25 extra-hard Quant Tests

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Manager
Manager
User avatar
Joined: 07 May 2012
Posts: 76
Location: United States
Followers: 1

Kudos [?]: 45 [0], given: 23

Re: For every integer m from 1 to 100, inclusive, the mth term [#permalink] New post 04 May 2013, 05:06
Ousmane wrote:
Why should we consider this as an infinite sequence?



Lets say we have only the first term m=1 , then the sum of the series (since it is just one term) , would be -1/2
Now , as explained by Bunuel , lets suppose there are infinite terms , in which case the sum is -1/3 . Hence sum of 1 to 100 terms SHOULD lie between these two range "-1/2 and -1/3".

-Jyothi
_________________

Jyothi hosamani

Intern
Intern
User avatar
Joined: 06 Jan 2013
Posts: 24
GPA: 3
WE: Engineering (Transportation)
Followers: 0

Kudos [?]: 21 [0], given: 9

GMAT ToolKit User
Re: For every integer m from 1 to 100, inclusive, the mth term [#permalink] New post 05 May 2013, 08:45
A general term of the expression is:-
(-1)^m.2^{-m}
Hence, the 1st 100 terms can be written as:-
t_1=\frac{-1}{2}
t_2=\frac{1}{4}
t_3=\frac{-1}{8}
.
.
.
.
.
t_{100}=\frac{1}{2^{100}}
Now, we know that the sum of the first "n" terms of a Geometric progression with a common ratio r<1 is:
\frac{a(1-r^n)}{1-r}
where a=1st term, r=common ratio and n=no. of terms.
Now for this case the expression would be(after putting all the values):-
N=\frac{\frac{-1}{2}[1-(\frac{-1}{2})^{100}]}{1-(\frac{-1}{2})}
=\frac{\frac{-1}{2}[1-\frac{1}{2^{100}}]}{\frac{3}{2}}
=\frac{-1}{3}.[1-\frac{1}{2^{100}}]
Now, since 2^{100} is a very large number compared to 1,
Hence, we can cay that 1-(\frac{1}{2^{100}}) is slightly less than 1
Hence, N is slightly less than \frac{-1}{3}.
_________________

If you shut your door to all errors, truth will be shut out.

CEO
CEO
User avatar
Joined: 09 Sep 2013
Posts: 2934
Followers: 211

Kudos [?]: 44 [0], given: 0

Premium Member
Re: For every integer m from 1 to 100, inclusive, the mth term [#permalink] New post 17 Jul 2014, 09:24
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

GMAT Books | GMAT Club Tests | Best Prices on GMAT Courses | GMAT Mobile App | Math Resources | Verbal Resources

Re: For every integer m from 1 to 100, inclusive, the mth term   [#permalink] 17 Jul 2014, 09:24
    Similar topics Author Replies Last post
Similar
Topics:
For every integer k from 1 to 10, inclusive, the kth term... njkhokh 0 04 Aug 2013, 11:05
1) For every integer K from 1 to 10 inclusive, the Kth term zakk 6 11 Aug 2007, 12:14
For every integer K from 1-10 inclusive, the kth term of a shehreenquayyum 1 05 Nov 2006, 13:08
1 For every integer k from 1 to 10 inclusive the kth term of a kuristar 3 11 Jun 2006, 16:04
For every integer k from 1 to 10, inclusive, the kth term of macca 5 27 Mar 2006, 13:56
Display posts from previous: Sort by

For every integer m from 1 to 100, inclusive, the mth term

  Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.