Summer is Coming! Join the Game of Timers Competition to Win Epic Prizes. Registration is Open. Game starts Mon July 1st.

It is currently 18 Jul 2019, 15:23

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.

Close

Request Expert Reply

Confirm Cancel

For integers n from 1 to 11, inclusive, the nth term of a ce

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Find Similar Topics 
Intern
Intern
avatar
Joined: 01 Aug 2013
Posts: 11
Location: India
For integers n from 1 to 11, inclusive, the nth term of a ce  [#permalink]

Show Tags

New post 14 Oct 2013, 09:37
1
6
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

67% (02:04) correct 33% (02:27) wrong based on 233 sessions

HideShow timer Statistics


For integers n from 1 to 11, inclusive, the nth term of a certain sequence is given by
Attachment:
Q.png
Q.png [ 1.35 KiB | Viewed 4438 times ]
.
If S is the sum of the first 11 terms in the sequence then S is

(A) Greater than 2

(B) Between 1 and 2

(C) Between 0 and 1/2

(D) Between -1/2 and 0

(E) Less than -1/2

I got the answer correct but i am looking for the easiest and fastest method to solve this.
Most Helpful Community Reply
Verbal Forum Moderator
User avatar
B
Joined: 10 Oct 2012
Posts: 604
Re: For integers n from 1 to 11, inclusive, the nth term of a ce  [#permalink]

Show Tags

New post 14 Oct 2013, 10:08
5
2
praffulpatel wrote:
For integers n from 1 to 11, inclusive, the nth term of a certain sequence is given by
Attachment:
Q.png
.
If S is the sum of the first 11 terms in the sequence then S is

(A) Greater than 2

(B) Between 1 and 2

(C) Between 0 and 1/2

(D) Between -1/2 and 0

(E) Less than -1/2

I got the answer correct but i am looking for the easiest and fastest method to solve this.


Method I :Crude

The fastest way is to realize that

1. The sign of the term with the greatest absolute value( that is the first term) has a negative sign attached to it
2. The given range is all within the precision of 1 decimal value,i.e. between 0.0 and -0.5 or less than -0.5
3.The greater the denominator, the more will be the no of decimal places

Take the sum of the first two terms : \(-\frac{1}{3}+\frac{1}{9} = \frac{1-3}{9} = \frac{-2}{9} = -2*0.11(repeating) = -0.222222....\)
Rest of the remaining terms, can ONLY change the above value after the first decimal place. Thus, the value would hover around -0.2XXXX and thus the answer is D.

Method II :

A little observation and fair pairing of the terms would show that the sum of the first two terms : \(\frac{-2}{3^2}\) , the pair of the 3rd and 4th terms : \(\frac{-2}{3^4}\) and so on.

Take an infinite sum of the series with terms as \(\frac{-2}{3^2} , \frac{-2}{3^4},.. \to -2*(\frac{1}{3^2}+\frac{1}{3^4}+...) = -2*\frac{\frac{1}{3^2}}{1-\frac{1}{9}\) = \(-2*\frac{1}{8} = -0.25\). Thus, the given series can never sum up-to a value which is more than -0.25.

D.
_________________
General Discussion
Intern
Intern
avatar
Joined: 05 Oct 2013
Posts: 21
Re: For integers n from 1 to 11, inclusive, the nth term of a ce  [#permalink]

Show Tags

New post 26 Oct 2013, 07:24
praffulpatel wrote:
For integers n from 1 to 11, inclusive, the nth term of a certain sequence is given by
Attachment:
Q.png
.
If S is the sum of the first 11 terms in the sequence then S is

(A) Greater than 2

(B) Between 1 and 2

(C) Between 0 and 1/2

(D) Between -1/2 and 0

(E) Less than -1/2

I got the answer correct but i am looking for the easiest and fastest method to solve this.


We have \(S = \sum_{n=1}^{11} \frac{(-1)^n}{3^n} =\frac{-1}{3} \sum_{n=0}^{10} \frac{(-1)^n}{3^n} = \frac{-1}{3} \frac{ 1 - \frac{(-1)^{11}}{3^{11}}}{1 - \frac{-1}{3}}\). So S is approximately equal to \(\frac{-1}{3} \frac{ 1 }{\frac{4}{3}} = \frac{-1}{4}\).
Therefore, the answer is D
Intern
Intern
avatar
Joined: 03 Oct 2013
Posts: 6
Re: For integers n from 1 to 11, inclusive, the nth term of a ce  [#permalink]

Show Tags

New post 28 Oct 2013, 15:03
2
The series is -1/3+1/3^2...So the value of this series has to be higher than -1/3
Further, this is a GP with r=-1/3
Assuming this goes on till infinity the sum would equal a/(1-r) = -1/3/(1-(-1/3))=-1/4
hence the sum of this series has to lie between -1/3 and -1/4
Intern
Intern
avatar
B
Status: Miles to go....before i sleep
Joined: 26 May 2013
Posts: 14
Location: India
Concentration: Finance, Marketing
Schools: ESADE '21 (A)
GMAT 1: 670 Q47 V35
Reviews Badge
Re: For integers n from 1 to 11, inclusive, the nth term of a ce  [#permalink]

Show Tags

New post 05 Dec 2013, 11:01
1
3
The sequence is given by (-1)^n*(1/3^n)
Putting n=1, the above sequence gives us the value, -1/3
Putting n=2, the above sequence gives us the value 1/3^2
Putting n=3, the above sequence gives us the value -1/3^3
So it forms a sequence -1/3,1/3^2,-1/3^3,1/3^4....
The above sequence will turn out to be in Geometric Progression(GP) with common ration (r) = -1/3

The easiest way to solve is by following the below approach which says that.
If the absolute value of common ratio |r|<1, then the Sum of GP is defined by
Sum=\(\frac{a}{1-r}\) , where a= First term of the GP, and r = common ratio


As we can see the 'r' in our sequence is -1/3, hence |r|<1, so the sum of GP can be given as
\(S =\frac{-1}{3}/1-(\frac{-1}{3}) => \frac{-1}{4}\)

Now look out the answer choices.
Answer D is the correct answer since -1/4 is between -1/2 and 0
Director
Director
User avatar
G
Joined: 23 Jan 2013
Posts: 542
Schools: Cambridge'16
For integers n from 1 to 11, inclusive, the nth term of a ce  [#permalink]

Show Tags

New post 06 Sep 2015, 21:56
First term=-1^1*(1/3)=-1/3
Second term=-1^2*(1/9)=1/9,
Sum=-1/3+1/9=-2/9, so every pair (3, 4th; 5,6th) sum will be negative. Total sum will be negative and in -1/3<S<0

D
Intern
Intern
avatar
B
Joined: 01 Sep 2018
Posts: 35
GMAT 1: 710 Q48 V40
GPA: 3.5
WE: Information Technology (Insurance)
Re: For integers n from 1 to 11, inclusive, the nth term of a ce  [#permalink]

Show Tags

New post 06 Jan 2019, 19:09
Bunuel

Is the following a good approach:

Step 1: Realize that there are 11 terms in total --> means that -1^n will stay negative. This gets rid off A, B, and C.
Step 2: n1 is -(1/3)
Step 3: n11 is - (1/(3^11)) --> probably very close to 0. Dont know 3^11 so not even going to calculate.
Step 4: other values will fall in between, so this means that D is the right choice
GMAT Club Bot
Re: For integers n from 1 to 11, inclusive, the nth term of a ce   [#permalink] 06 Jan 2019, 19:09
Display posts from previous: Sort by

For integers n from 1 to 11, inclusive, the nth term of a ce

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  





Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne