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 350,000 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: PS Question 700 level - Need easy to solve this [#permalink]
19 Feb 2010, 13:14

2

This post received KUDOS

Expert's post

shuj00 wrote:

The infinite sequence a1, a2,...,an is defined such that an = (n+2) / n for all n ≥ 1. What is the product of the first 10 terms of the sequence?

(A) 45 (B) 66 (C) 90 (D) 121 (E) 132

It's very often a good idea in sequence questions to write down the first few terms; typically a pattern will emerge. Here we find each term by plugging n into the expression given:

Notice that almost the entire denominator can be canceled, leaving us with (11*12)/2 = 66. _________________

GMAT Tutor in Toronto

If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com

Re: Sum of consecutive terms [#permalink]
15 Nov 2010, 17:56

1

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

Yalephd wrote:

The infinite sequence a1, a2,...,an is defined such that an = (n+2) / n for all n ≥ 1. What is the product of the first 10 terms of the sequence?

(A) 45 (B) 66 (C) 90 (D) 121 (E) 132

Powering through this would take the full lenght of the GMAT (maybe not). Is there a quick way to solve this problem?

Look at the pattern: \(an = \frac{(n + 2)}{n}\) So a1 = 3/1 a2 = 4/2 a3 = 5/3 a4 = 6/4

Now when we multiply the first ten terms (n = 1 to 10), \(\frac{3}{1} * \frac{4}{2} * \frac{5}{3} * \frac{6}{4} * ...* \frac{11}{9} * \frac{12}{10}\) denominators and numerators will get canceled, 3 with 3, 4 with 4 etc. Only the first two denominators and last two numerators will be left. The product will be 11*12/2 = 66 _________________

Re: Sum of consecutive terms [#permalink]
09 Dec 2010, 08:12

VeritasPrepKarishma wrote:

Yalephd wrote:

The infinite sequence a1, a2,...,an is defined such that an = (n+2) / n for all n ≥ 1. What is the product of the first 10 terms of the sequence?

(A) 45 (B) 66 (C) 90 (D) 121 (E) 132

Powering through this would take the full lenght of the GMAT (maybe not). Is there a quick way to solve this problem?

Look at the pattern: \(an = \frac{(n + 2)}{n}\) So a1 = 3/1 a2 = 4/2 a3 = 5/3 a4 = 6/4

Now when we multiply the first ten terms (n = 1 to 10), \(\frac{3}{1} * \frac{4}{2} * \frac{5}{3} * \frac{6}{4} * ...* \frac{11}{9} * \frac{12}{10}\) denominators and numerators will get canceled, 3 with 3, 4 with 4 etc. Only the first two denominators and last two numerators will be left. The product will be 11*12/2 = 66

Since this is a sequence and seeing the pattern " increase of numerator and denominator by 1/1" we determine the next term.This part I understood. But fundamental thing that I have not yet grasped is how to visualise that the respective denominators and numerators actually cancel without enumerating them all. I had to list out all the fractions (just by adding 1/1 to the next) to see that they do cancel and what the final fractions would remain (though in matters of time I did not consume much time-few extra seconds). Is there a logic behind this visualisation ? Please advice.

Re: Sum of consecutive terms [#permalink]
09 Dec 2010, 11:04

1

This post received KUDOS

Expert's post

helloanupam wrote:

VeritasPrepKarishma wrote:

Yalephd wrote:

The infinite sequence a1, a2,...,an is defined such that an = (n+2) / n for all n ≥ 1. What is the product of the first 10 terms of the sequence?

(A) 45 (B) 66 (C) 90 (D) 121 (E) 132

Powering through this would take the full lenght of the GMAT (maybe not). Is there a quick way to solve this problem?

Look at the pattern: \(an = \frac{(n + 2)}{n}\) So a1 = 3/1 a2 = 4/2 a3 = 5/3 a4 = 6/4

Now when we multiply the first ten terms (n = 1 to 10), \(\frac{3}{1} * \frac{4}{2} * \frac{5}{3} * \frac{6}{4} * ...* \frac{11}{9} * \frac{12}{10}\) denominators and numerators will get canceled, 3 with 3, 4 with 4 etc. Only the first two denominators and last two numerators will be left. The product will be 11*12/2 = 66

Since this is a sequence and seeing the pattern " increase of numerator and denominator by 1/1" we determine the next term.This part I understood. But fundamental thing that I have not yet grasped is how to visualise that the respective denominators and numerators actually cancel without enumerating them all. I had to list out all the fractions (just by adding 1/1 to the next) to see that they do cancel and what the final fractions would remain (though in matters of time I did not consume much time-few extra seconds). Is there a logic behind this visualisation ? Please advice.

The nth term is (n+2)/n.

So I say, when I multiply terms, they will look like this: \(\frac{[highlight](n+2)[/highlight]}{n} * \frac{(n+3)}{(n+1)} * \frac{(n+4)}{(n+2)} * \frac{(n+5)}{(n+3)} * ... * \frac{(n+12)}{(n+10)}\) Watch what happens here. The numerator (n+2) of the 1st term has appeared in the denominator of the 3rd term so they will cancel out. So 2nd term will have numerator that is 1 more than the numerator of the 1st term and 4th term will have a denominator that is one more than the denominator of the 3rd term and hence they will be equal and will cancel out. This will continue till we run out of denominators to cancel. So even without enumerating, we know that we will be left with first two numerators and last two denominators.

(I wish I could strike the relevant terms with different colored lines but I haven't been able to figure out how to use other operators in fractions!) _________________

Re: Sum of consecutive terms [#permalink]
09 Dec 2010, 23:56

Expert's post

gettinit wrote:

How do we know a1=1? we don't know the value of the first 10 terms? how can we assume a1=1? I didn't see that and was confused by the notation.

\(a_1\) doesn't equal to 1, it equals to \(a_1=\frac{1+2}{1}=\frac{3}{1}\).

The infinite sequence a1, a2,..., an is defined such that an = (n+2) / n for all n ≥ 1. What is the product of the first 10 terms of the sequence? (A) 45 (B) 66 (C) 90 (D) 121 (E) 132

Stem gives the formula to get the terms of the sequences: \(a_n=\frac{n+2}{n}\) for ALL \(n\geq{1}\) (so starting from \(a_1\)). So:

The product of the first 10 terms of the sequence will equal: \(\frac{3}{1}*\frac{4}{2}*\frac{5}{3}*...*\frac{12}{10}=\frac{(3*4*5*6*6*8*9*10)*11*12}{1*2*(3*4*5*6*6*8*9*10)}=\frac{11*12}{2}=66\).

Re: Sum of consecutive terms [#permalink]
10 Dec 2010, 04:39

Thanks Karishma. I took a couple of bounces before I understood from your explanation albeit in own fashion (and hopefully correct!).The way I understand is this- the first two denominators (n & n+1) do not have their mirror numerators and similarly the final two numerators (n+11) & (n+12) wont get to see their mirror denominator images as the process stops at denominator (n+10) and hence dont cancel out.

Re: Sum of consecutive terms [#permalink]
10 Dec 2010, 04:52

Expert's post

gettinit wrote:

How do we know a1=1? we don't know the value of the first 10 terms? how can we assume a1=1? I didn't see that and was confused by the notation.

This is what is given to you in the question: The infinite sequence a1, a2,...,an is defined such that an = (n+2) / n for all n ≥ 1.

The product of first 10 terms is asked.

Using the rule, an = (n+2) / n, you can find the nth term easily. The first term will have n = 1 a1 = (1+2)/1 = 3/1 a2 = (2+2)/2 = 4/2 and so on... _________________

Re: Sum of consecutive terms [#permalink]
10 Dec 2010, 04:54

Expert's post

helloanupam wrote:

Thanks Karishma. I took a couple of bounces before I understood from your explanation albeit in own fashion (and hopefully correct!).The way I understand is this- the first two denominators (n & n+1) do not have their mirror numerators and similarly the final two numerators (n+11) & (n+12) wont get to see their mirror denominator images as the process stops at denominator (n+10) and hence dont cancel out.

The infinite sequence a1, a2,...,an is defined such that an = (n [#permalink]
26 Dec 2012, 18:49

shuj00 wrote:

The infinite sequence a1, a2,...,an is defined such that an = (n+2) / n for all n ≥ 1. What is the product of the first 10 terms of the sequence?

(A) 45 (B) 66 (C) 90 (D) 121 (E) 132

Seeing that the series is a sequence of fractions. Seeing that the question is asking about the product... we know that factors will cancel out and easily reveal the answer.

Re: The infinite sequence a1, a2,...,an is defined such that an [#permalink]
05 Jul 2014, 23:32

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. _________________

Re: The infinite sequence a1, a2,...,an is defined such that an [#permalink]
16 Jul 2015, 18:13

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. _________________

The Stanford interview is an alumni-run interview. You give Stanford your current address and they reach out to alumni in your area to find one that can interview you...

Originally, I was supposed to have an in-person interview for Yale in New Haven, CT. However, as I mentioned in my last post about how to prepare for b-school interviews...

Interested in applying for an MBA? In the fourth and final part of our live QA series with guest expert Chioma Isiadinso, co-founder of consultancy Expartus and former admissions...