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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

Show Tags

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?

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

Show Tags

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 !! Thanks

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

Show Tags

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

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.

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

Show Tags

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.
_________________

We’ve given one of our favorite features a boost! You can now manage your profile photo, or avatar , right on WordPress.com. This avatar, powered by a service...

Sometimes it’s the extra touches that make all the difference; on your website, that’s the photos and video that give your content life. You asked for streamlined access...

A lot has been written recently about the big five technology giants (Microsoft, Google, Amazon, Apple, and Facebook) that dominate the technology sector. There are fears about the...

Post today is short and sweet for my MBA batchmates! We survived Foundations term, and tomorrow's the start of our Term 1! I'm sharing my pre-MBA notes...