It is currently 19 Oct 2017, 11:58

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

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

For all n such that n is a positive integer, the terms of a certain

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

Hide Tags

Expert Post
1 KUDOS received
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 41892

Kudos [?]: 128890 [1], given: 12183

For all n such that n is a positive integer, the terms of a certain [#permalink]

Show Tags

New post 08 Jun 2015, 09:08
1
This post received
KUDOS
Expert's post
5
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

74% (01:41) correct 26% (02:36) wrong based on 106 sessions

HideShow timer Statistics

For all n such that n is a positive integer, the terms of a certain sequence B are given by the following rules:

\(B_n = B_{n-1} + 5\) if n is odd and greater than 1;
\(Bn = -B_{n-1}\) if n is even;
\(B_1 = 3\)

What is the sum of the first 65 terms in the sequence?

(A) –5
(B) 0
(C) 3
(D) 5
(E) 8

Kudos for a correct solution.
[Reveal] Spoiler: OA

_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | 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?
Extra-hard Quant Tests with Brilliant Analytics

Kudos [?]: 128890 [1], given: 12183

5 KUDOS received
Math Forum Moderator
User avatar
Joined: 06 Jul 2014
Posts: 1272

Kudos [?]: 2313 [5], given: 178

Location: Ukraine
Concentration: Entrepreneurship, Technology
GMAT 1: 660 Q48 V33
GMAT 2: 740 Q50 V40
GMAT ToolKit User Premium Member
Re: For all n such that n is a positive integer, the terms of a certain [#permalink]

Show Tags

New post 08 Jun 2015, 09:39
5
This post received
KUDOS
Bunuel wrote:
For all n such that n is a positive integer, the terms of a certain sequence B are given by the following rules:

\(B_n = B_{n-1} + 5\) if n is odd and greater than 1;
\(Bn = -B_{n-1}\) if n is even;
\(B_1 = 3\)

What is the sum of the first 65 terms in the sequence?

(A) –5
(B) 0
(C) 3
(D) 5
(E) 8

Kudos for a correct solution.


B1 = 3
B2 = -3
B3 = 2
B4 = -2
B5 = 3 and from this moment we see that pattern will be repeat each four numbers
65 / 4 = 4*16 + 1

so B65 will be equal to 3 (16 repetions + 1 more calculation)

Answer is C
_________________

Simple way to always control time during the quant part.
How to solve main idea questions without full understanding of RC.
660 (Q48, V33) - unpleasant surprise
740 (Q50, V40, IR3) - anti-debrief ;)

Kudos [?]: 2313 [5], given: 178

1 KUDOS received
Senior Manager
Senior Manager
avatar
Joined: 27 Dec 2013
Posts: 304

Kudos [?]: 37 [1], given: 113

Re: For all n such that n is a positive integer, the terms of a certain [#permalink]

Show Tags

New post 08 Jun 2015, 09:55
1
This post received
KUDOS
1
This post was
BOOKMARKED
The correct answer= C.

The first four terms in the sequence are 3, -3,+3,-3. Repeat teh sequence for 16 times. You get 64 numbers and the sum is 0. The last number will be 3.

KR.

Bunuel wrote:
For all n such that n is a positive integer, the terms of a certain sequence B are given by the following rules:

\(B_n = B_{n-1} + 5\) if n is odd and greater than 1;
\(Bn = -B_{n-1}\) if n is even;
\(B_1 = 3\)

What is the sum of the first 65 terms in the sequence?

(A) –5
(B) 0
(C) 3
(D) 5
(E) 8

Kudos for a correct solution.

_________________

Kudos to you, for helping me with some KUDOS.

Kudos [?]: 37 [1], given: 113

1 KUDOS received
Manager
Manager
User avatar
Joined: 26 Dec 2011
Posts: 121

Kudos [?]: 100 [1], given: 44

Schools: HBS '18, IIMA
Re: For all n such that n is a positive integer, the terms of a certain [#permalink]

Show Tags

New post 08 Jun 2015, 10:10
1
This post received
KUDOS
First four terms are repeating B1=3, B2=-3, B3=2, B4=-2....

There are 16 such repetitions equal to 64 and the 65th term is B65=B1=3.


Thanks
_________________

Thanks,
Kudos Please

Kudos [?]: 100 [1], given: 44

1 KUDOS received
Manager
Manager
User avatar
Joined: 21 Jan 2015
Posts: 150

Kudos [?]: 104 [1], given: 24

Location: India
Concentration: Strategy, Marketing
WE: Marketing (Other)
GMAT ToolKit User
Re: For all n such that n is a positive integer, the terms of a certain [#permalink]

Show Tags

New post 09 Jun 2015, 02:00
1
This post received
KUDOS
Bunuel wrote:
For all n such that n is a positive integer, the terms of a certain sequence B are given by the following rules:

\(B_n = B_{n-1} + 5\) if n is odd and greater than 1;
\(Bn = -B_{n-1}\) if n is even;
\(B_1 = 3\)

What is the sum of the first 65 terms in the sequence?

(A) –5
(B) 0
(C) 3
(D) 5
(E) 8


Ans: C

Solution: by given condition we start from B1=3, B2=-3, B3=2, B4=-2, B5=3, B6=-3 B7=2, B8=-2....... and so on
so this sequence repeat first 4 terms with the cycle of 4 terms. and there sum is 0.
so now first 65 terms had 16 full cycle and one extra term B65=3

so the sum of first 64 terms is 3.
_________________

--------------------------------------------------------------------
The Mind is Everything, What we Think we Become.
Kudos will encourage many others, like me.
Please Give Kudos Image !!
Thanks :-)

Kudos [?]: 104 [1], given: 24

Current Student
avatar
Joined: 29 Mar 2015
Posts: 44

Kudos [?]: 12 [0], given: 9

Location: United States
GMAT ToolKit User Reviews Badge
Re: For all n such that n is a positive integer, the terms of a certain [#permalink]

Show Tags

New post 10 Jun 2015, 01:56
1
This post was
BOOKMARKED
Bn=Bn−1+5 if n is odd and greater than 1;
Bn=−Bn−1 if n is even;
B1=3
\(B_1 =3\)
\(B_2 =-B_1=-3\) because 2 is even
\(B_3 =B_2+5=2\)
\(B_4 =-B_3=-2\)
\(B_5 =B_4+5=3\)
\(B_6 =-B_5=-3\) etc.
so it is \(3-3+2-2+3-3+...\) 35 times, with every two numbers summing to 0. For the first 65 terms, there are 32 pairs and then the 65th term. The 65th term is 3, because it starts the 33th pair, and the start of every odd pair is 3. So the answer is C

Kudos [?]: 12 [0], given: 9

1 KUDOS received
Manager
Manager
avatar
Joined: 01 Jan 2015
Posts: 56

Kudos [?]: 3 [1], given: 7

Re: For all n such that n is a positive integer, the terms of a certain [#permalink]

Show Tags

New post 14 Jun 2015, 04:47
1
This post received
KUDOS
The first four terms are 3, -3,+3,-3.
sequence for 64 numbers = sum is 0.
Last number will be 3.

Answer= C?

Kudos [?]: 3 [1], given: 7

Expert Post
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 41892

Kudos [?]: 128890 [0], given: 12183

Re: For all n such that n is a positive integer, the terms of a certain [#permalink]

Show Tags

New post 15 Jun 2015, 03:45
Expert's post
1
This post was
BOOKMARKED
Bunuel wrote:
For all n such that n is a positive integer, the terms of a certain sequence B are given by the following rules:

\(B_n = B_{n-1} + 5\) if n is odd and greater than 1;
\(Bn = -B_{n-1}\) if n is even;
\(B_1 = 3\)

What is the sum of the first 65 terms in the sequence?

(A) –5
(B) 0
(C) 3
(D) 5
(E) 8

Kudos for a correct solution.


MANHATTAN GMAT OFFICIAL SOLUTION:

Let's list the first few terms of the sequence according to the given rules:
B1 = 3
B2 = -B1 = –3
B3 = B2 + 5 = –3 + 5 = 2
B4 = -B3 = –2
B5 = B4 + 5 = –2 + 5 = 3
…etc.

Note that the pattern is a four-term repeat: 3, –3 , 2, –2 . Also note that the sum of this repeating group is (3) + (–3) + (2) + (–2) = 0. This repeating group will occur 16 times through term number 64. Thus, the sum of the first 64 terms will be 0. This leaves the 65th term, which will have the same value as B1 : 3. Therefore, the sum of the first 65 terms is 3.

The correct answer is C.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | 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?
Extra-hard Quant Tests with Brilliant Analytics

Kudos [?]: 128890 [0], given: 12183

GMAT Club Legend
GMAT Club Legend
User avatar
Joined: 09 Sep 2013
Posts: 16654

Kudos [?]: 273 [0], given: 0

Premium Member
Re: For all n such that n is a positive integer, the terms of a certain [#permalink]

Show Tags

New post 04 Aug 2017, 13:54
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

Kudos [?]: 273 [0], given: 0

Re: For all n such that n is a positive integer, the terms of a certain   [#permalink] 04 Aug 2017, 13:54
Display posts from previous: Sort by

For all n such that n is a positive integer, the terms of a certain

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


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

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

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®.