It is currently 17 Feb 2019, 08:10
My Tests

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.

sequence of numbers a1, a2, a3, . . . , an

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
GRE Prep Club Legend
GRE Prep Club Legend
User avatar
Joined: 07 Jun 2014
Posts: 4856
GRE 1: Q167 V156
WE: Business Development (Energy and Utilities)
Followers: 102

Kudos [?]: 1736 [0], given: 397

sequence of numbers a1, a2, a3, . . . , an [#permalink] New post 04 Jan 2016, 06:20
Expert's post
00:00

Question Stats:

60% (01:26) correct 40% (01:40) wrong based on 15 sessions
The sequence of numbers a1, a2, a3, . . . , an, . . . is defined by \(an= \frac{1}{n} - \frac{1}{(n+2)}\) for each integer n ≥ 1. What is the sum of the first 20 terms of this sequence?

A. \((1+\frac{1}{2}) - \frac{1}{20}\)
B. \((1+\frac{1}{2}) - (\frac{1}{21}+\frac{1}{22})\)
C. \(1 - (\frac{1}{20}+\frac{1}{22})\)
D. \(1 - \frac{1}{22}\)
E. \(\frac{1}{20} - \frac{1}{22}\)





Practice Questions
Question: 12
Page: 158
Difficulty: hard
[Reveal] Spoiler: OA

_________________

Sandy
If you found this post useful, please let me know by pressing the Kudos Button

Try our free Online GRE Test

GRE Prep Club Legend
GRE Prep Club Legend
User avatar
Joined: 07 Jun 2014
Posts: 4856
GRE 1: Q167 V156
WE: Business Development (Energy and Utilities)
Followers: 102

Kudos [?]: 1736 [0], given: 397

Re: sequence of numbers a1, a2, a3, . . . , an [#permalink] New post 04 Jan 2016, 08:40
Expert's post
Here we have \(an= \frac{1}{n} - \frac{1}{(n+2)}\). The idea is to find a pattern in the sum

\(a1= \frac{1}{1} - \frac{1}{3}\)
\(a2= \frac{1}{2} - \frac{1}{4}\)
\(a3= \frac{1}{3} - \frac{1}{5}\)
\(a4= \frac{1}{4} - \frac{1}{6}\)

Now we have the first term from a3 can cancels second term from previous numbers

Now observing the last few terms...

\(a18= \frac{1}{18} - \frac{1}{20}\)
\(a19= \frac{1}{19} - \frac{1}{21}\)
\(a20= \frac{1}{20} - \frac{1}{22}\)

Now we have here the last two terms namely \(\frac{1}{21}\) and \(\frac{1}{22}\) are not cancelled.

\((1+\frac{1}{2}) - (\frac{1}{21}+\frac{1}{22})\)

Hence the sum is
_________________

Sandy
If you found this post useful, please let me know by pressing the Kudos Button

Try our free Online GRE Test

Intern
Intern
Joined: 28 Mar 2016
Posts: 11
Followers: 0

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

Re: sequence of numbers a1, a2, a3, . . . , an [#permalink] New post 29 Mar 2016, 09:34
Now we have the first term from a3 can cancels second term from previous numbers
^^^ what do you mean by this exactly, like what exact numbers cancel?

does the sum of the first two cancel, or something else?
1 KUDOS received
Intern
Intern
Joined: 12 Jul 2016
Posts: 8
Followers: 0

Kudos [?]: 4 [1] , given: 11

Re: sequence of numbers a1, a2, a3, . . . , an [#permalink] New post 13 Jul 2016, 13:31
1
This post received
KUDOS
sagnik242 wrote:
Now we have the first term from a3 can cancels second term from previous numbers
^^^ what do you mean by this exactly, like what exact numbers cancel?

does the sum of the first two cancel, or something else?



In the first set, you cancel both the 1/3, the 1/4, and the 1/5. You see the pattern now? This cancellation will continue until the last two terms (1/21 and 1/20 cannot be cancelled because n=20 is the limit).

In the first set you are left with 1/1 and 1/2. By the second set there's only 1/21 and 1/22. You add the first set and subtract the second. (1/1 + 1/2) - (1/21 + 1/22).
2 KUDOS received
GRE Instructor
User avatar
Joined: 10 Apr 2015
Posts: 1408
Followers: 54

Kudos [?]: 1347 [2] , given: 8

CAT Tests
Re: sequence of numbers a1, a2, a3, . . . , an [#permalink] New post 09 Feb 2019, 08:17
2
This post received
KUDOS
Expert's post
sandy wrote:
The sequence of numbers a1, a2, a3, . . . , an, . . . is defined by \(an= \frac{1}{n} - \frac{1}{(n+2)}\) for each integer n ≥ 1. What is the sum of the first 20 terms of this sequence?

A. \((1+\frac{1}{2}) - \frac{1}{20}\)
B. \((1+\frac{1}{2}) - (\frac{1}{21}+\frac{1}{22})\)
C. \(1 - (\frac{1}{20}+\frac{1}{22})\)
D. \(1 - \frac{1}{22}\)
E. \(\frac{1}{20} - \frac{1}{22}\)



Applying the formula, we get:
a1 = (1/1 - 1/3)
a2 = (1/2 - 1/4)
a3 = (1/3 - 1/5)
a4 = (1/4 - 1/6)
.
.
.
.
a17 = (1/17 - 1/19)
a18 = (1/18 - 1/20)
a19 = (1/19 - 1/21)
a20 = (1/20 - 1/22)

So, the SUM = (1/1 - 1/3) + (1/2 - 1/4) + (1/3 - 1/5) + (1/4 - 1/6) . . . (1/17 - 1/19) + (1/18 - 1/20) + (1/19 - 1/21) + (1/20 - 1/22)

Notice that all of the SAME COLORED fractions cancel out.
For example, (-1/3) + 1/3 = 0

The only fractions the DON'T get canceled out are those at the beginning and end of the sequence.
So, SUM = 1/1 + 1/2 - 1/21 - 1/22
We can rewrite this as: SUM = (1/1 + 1/2) - (1/21 + 1/22)

Answer: B

Cheers,
Brent
_________________

Brent Hanneson – Creator of greenlighttestprep.com
Image
Sign up for our free GRE Question of the Day emails

Re: sequence of numbers a1, a2, a3, . . . , an   [#permalink] 09 Feb 2019, 08:17
Display posts from previous: Sort by

sequence of numbers a1, a2, a3, . . . , an

  Question banks Downloads My Bookmarks Reviews Important topics  


cron

GRE Prep Club Forum Home| About| Terms and Conditions and Privacy Policy| GRE Prep Club Rules| Contact

Powered by phpBB © phpBB Group

Kindly note that the GRE® test is a registered trademark of the Educational Testing Service®, and this site has neither been reviewed nor endorsed by ETS®.