GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 19 Nov 2018, 03:31

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 in November
PrevNext
SuMoTuWeThFrSa
28293031123
45678910
11121314151617
18192021222324
2526272829301
Open Detailed Calendar
  • How to QUICKLY Solve GMAT Questions - GMAT Club Chat

     November 20, 2018

     November 20, 2018

     09:00 AM PST

     10:00 AM PST

    The reward for signing up with the registration form and attending the chat is: 6 free examPAL quizzes to practice your new skills after the chat.
  • The winning strategy for 700+ on the GMAT

     November 20, 2018

     November 20, 2018

     06:00 PM EST

     07:00 PM EST

    What people who reach the high 700's do differently? We're going to share insights, tips and strategies from data we collected on over 50,000 students who used examPAL.

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

Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 50661
For all n such that n is a positive integer, the terms of a certain  [#permalink]

Show Tags

New post 08 Jun 2015, 08:08
3
5
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

75% (02:34) correct 25% (02:31) wrong based on 181 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.

_________________

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

Most Helpful Community Reply
Retired Moderator
User avatar
Joined: 06 Jul 2014
Posts: 1241
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, 08:39
5
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 ;)

General Discussion
Manager
Manager
avatar
Joined: 27 Dec 2013
Posts: 249
Re: For all n such that n is a positive integer, the terms of a certain  [#permalink]

Show Tags

New post 08 Jun 2015, 08:55
1
1
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.

Manager
Manager
User avatar
Joined: 26 Dec 2011
Posts: 114
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, 09:10
1
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

Senior Manager
Senior Manager
User avatar
S
Joined: 21 Jan 2015
Posts: 346
Location: India
Concentration: Strategy, Marketing
GMAT 1: 620 Q48 V28
GMAT 2: 690 Q49 V35
WE: Sales (Consumer Products)
GMAT ToolKit User CAT Tests
Re: For all n such that n is a positive integer, the terms of a certain  [#permalink]

Show Tags

New post 09 Jun 2015, 01:00
1
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 :-)

Current Student
avatar
Joined: 29 Mar 2015
Posts: 44
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, 00:56
1
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
Manager
Manager
avatar
Joined: 01 Jan 2015
Posts: 56
Re: For all n such that n is a positive integer, the terms of a certain  [#permalink]

Show Tags

New post 14 Jun 2015, 03:47
1
The first four terms are 3, -3,+3,-3.
sequence for 64 numbers = sum is 0.
Last number will be 3.

Answer= C?
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 50661
Re: For all n such that n is a positive integer, the terms of a certain  [#permalink]

Show Tags

New post 15 Jun 2015, 02:45
1
1
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

Senior Manager
Senior Manager
avatar
P
Joined: 26 Jun 2017
Posts: 369
Location: Russian Federation
Concentration: General Management, Strategy
WE: Information Technology (Other)
Re: For all n such that n is a positive integer, the terms of a certain  [#permalink]

Show Tags

New post 08 Sep 2018, 01:42
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.


This question would be a little more difficult iw were "2" in options.
VP
VP
User avatar
P
Joined: 09 Mar 2016
Posts: 1093
For all n such that n is a positive integer, the terms of a certain  [#permalink]

Show Tags

New post 08 Sep 2018, 05:12
Harley1980 wrote:
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


how do we get these values ?

B2 = -3
B3 = 2
B4 = -2

hi pushpitkc can you explain pls
VP
VP
User avatar
P
Joined: 09 Mar 2016
Posts: 1093
Re: For all n such that n is a positive integer, the terms of a certain  [#permalink]

Show Tags

New post 08 Sep 2018, 08:07
Bunuel wrote:
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.



Bunuel how can B2 be equal to -B1 = –3 ? :? can you explain the logic pls :)
Senior SC Moderator
avatar
V
Joined: 22 May 2016
Posts: 2111
Premium Member CAT Tests
For all n such that n is a positive integer, the terms of a certain  [#permalink]

Show Tags

New post 08 Sep 2018, 18:46
1
dave13 wrote:
Harley1980 wrote:
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


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

how do we get these values ?

B2 = -3
B3 = 2
B4 = -2

hi pushpitkc can you explain pls

dave13 , you will have to deal with my pinch-hitting for pushpitkc :)

I can't quite tell what part does not make sense, so I may give you too much info. The way that the instructions are written may be throwing you off.

This problem is like a recipe for a sequence. The "recipe" tells us how to calculate each term.

The "ingredients" are terms with even- and odd-numbered subscripts. Even- and odd-numbered terms have different rules.

We specify each term by using a subscript, \(B_{n}\). The first term is \(B_1\), where \(n=1\). Second term is \(B_2\), where \(n=2\)

As mentioned, if the subscript is an odd number, we follow one rule. If subscript is even, we follow another rule.

Rule (1): IF \(n\) in \(B_{n}\) is ODD and greater than 1, use this rule: \(B_{n}=B_{(n-1)}+5\)

That rule means: to calculate the value of THIS term, use the value of the preceding term and add 5 to it.
\((n-1)\) means "use the term with a subscript that is 1 less than this term's subscript"

If we were calculating \(B_5\), we would plug in the value of \(B_4\).(then add 5) \(B_5\):

\(B_{(n-1)}=B_{(5-1)}=B_4\)
\(B_5\) thus will equal \(B_4+5\)

Rule (2) IF \(n\) in \(B_{n}\) is EVEN, use this rule: \(B_{n} = -(B_{(n-1)})\)
Meaning: take the value of the preceding term (n-1), and negate it.

1) Start calculating sequence terms. We need a pattern. Given:
\(B_1=3\)
Next term, \(B_2\)? Use rule for evens. Negate the preceding term \(B_2=-(B_{(n-1)})=-(B_{(2-1)})=-(B_1)=-(3)\)
That is, \(B_2=-3\)
\(B_3\)? Use the rule for odds

\(B_1=3\)
\(B_2=-(B_{(2-1)})=-(B_1)=-3\)
\(B_3=(B_{(n-1)}+5)=(B_2+5)=(-3+5)=2\)
\(B_4=-(B_3)=-(2)=-2\)
\(B_5=(B_4+5)=(-2+5)=3\)
\(B_6=-(B_5)=-(3)=-3\)
\(B_7=(B_6+5)=(-3+5)=2\)
\(B_8=-(B_7)=-(2)=-2\)

We have a pattern that repeats in cycles of 4 terms: \(3, -3, 2, -2\)

2) Cyclicity
We need the sum of "the first 65 terms." We will use the cycles of four terms to find that sum. Cyclicity in this case is like integer powers and units digits cyclicity.

We need to know where, in the cycle of four terms, \(B_{65}\) will "fall."

To find where \(B_{65}\) falls in the cycle of four terms: Divide the subscript "65" by the cyclicity of "4"
Remainder? Match the remainder to a subscript. The term whose subscript equals the remainder has the same value as \(B_{65}\)

No remainder? Then \(B_{65}\) has the same value as \(B_4\) (the term is a multiple of 4)

\(\frac{65}{4}=16\) + remainder 1
Thus \(B_{65}=B_1=3\)

3) SUM??
If we were literally to sum all the terms, the first 8 would be
\(3+(-3)+2+(-2)+(3)+(-3)+2+(-2)=\)
No thanks. We do not need to sum all the terms.
Every 4 terms, the sum of terms = 0: \(3+(-3)+2+(-2)=0\)

\(B_{65}\) had a "remainder 1." One term before it, \(B_{64}\) was the last term in a cycle of four numbers

So up to the 64\(^{th}\) term, from \(B_1\) to \(B_{64}\), the sum is 0. (16 cycles of four terms; every 4 terms= 0).
\(B_{65}=3\)

\((0+0+0+0+0+0. . . +3)=3\)
The sum of the first 65 terms \(= 3\)
Hope that helps.:-)
GMAT Club Bot
For all n such that n is a positive integer, the terms of a certain &nbs [#permalink] 08 Sep 2018, 18:46
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  


cron
Copyright

GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| 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®.