Last visit was: 23 Apr 2026, 02:14 It is currently 23 Apr 2026, 02:14
Home
Messages
Tests
Schools
Events
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.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
805+ (Hard)|   Sequences|      
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 23 Apr 2026
Posts: 109,763
Own Kudos:
810,725
 [9]
Given Kudos: 105,850
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 109,763
Kudos: 810,725
 [9]
The nth term of a sequence a1, a2, a3 an is given by an = 1/(n(n+2)) 23 Nov 2023, 02:30
1
Kudos
Add Kudos
8
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

95% (hard)

Question Stats:

37% (03:04) correct 63% (02:19) wrong based on 59 sessions

History

Date
Time
Result
Not Attempted Yet
The \(n_{th}\) term of a sequence \(a_1, \ a_2, \ a_3 \ … \ a_n\) is given by \(a_n = \frac{1}{n(n+2)}\). What is the sum of the first 20 terms of the sequence?

(A) 300/440

(B) 325/462

(C) 303/462

(D) 375/450

(E) 650/462

ShowHide Answer
Official Answer
B
User avatar
sj296
User avatar
Joined: 09 Jul 2014
Last visit: 19 Mar 2025
Posts: 305
Own Kudos:
397
Given Kudos: 346
Location: India
Concentration: General Management, Finance
Schools: ISB '24
Schools: ISB '24
Posts: 305
Kudos: 397
The nth term of a sequence a1, a2, a3 an is given by an = 1/(n(n+2)) 24 Nov 2023, 10:34
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Tn = \(\frac{1}{n(n+2)}\) = \(\frac{[m][fraction]1}{n} \)- \(\frac{1}{(n+2)}\)/2[/fraction][/m]

2Sn = \(\frac{1}{1.3}\) + \(\frac{1}{2.4}\)+ \(\frac{1}{3.5}\)...........................\(\frac{1}{18.20}\) + \(\frac{1}{19.21}\) + \(\frac{1}{20.22}\)

2Sn = \(\frac{1}{1}\) - \(\frac{1}{3}\) + \(\frac{1}{2}\) - \(\frac{1}{4}\) + \(\frac{1}{3}\) - \(\frac{1}{5}\) ........................\(\frac{1}{18}\) - \(\frac{1}{20}\) + \(\frac{1}{19}\)- \(\frac{1}{21}\) + \(\frac{1}{20}\) - \(\frac{1}{22}\)

2Sn = 1 + \(\frac{1}{2}\) - \(\frac{1}{21}\) - \(\frac{1}{22}\) = \(\frac{650}{462}\)

Sn = \(\frac{325}{462}\)

O\(\)ption B
User avatar
abishekcmouli
User avatar
Joined: 20 Mar 2023
Last visit: 14 May 2025
Posts: 11
Own Kudos:
6
 [1]
Given Kudos: 43
Posts: 11
Kudos: 6
 [1]
The nth term of a sequence a1, a2, a3 an is given by an = 1/(n(n+2)) 28 Nov 2023, 07:51
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
sj296
Tn = \(\frac{1}{n(n+2)}\) = \(\frac{[m][fraction]1}{n} \)- \(\frac{1}{(n+2)}\)/2[/fraction][/m]

2Sn = \(\frac{1}{1.3}\) + \(\frac{1}{2.4}\)+ \(\frac{1}{3.5}\)...........................\(\frac{1}{18.20}\) + \(\frac{1}{19.21}\) + \(\frac{1}{20.22}\)

2Sn = \(\frac{1}{1}\) - \(\frac{1}{3}\) + \(\frac{1}{2}\) - \(\frac{1}{4}\) + \(\frac{1}{3}\) - \(\frac{1}{5}\) ........................\(\frac{1}{18}\) - \(\frac{1}{20}\) + \(\frac{1}{19}\)- \(\frac{1}{21}\) + \(\frac{1}{20}\) - \(\frac{1}{22}\)

2Sn = 1 + \(\frac{1}{2}\) - \(\frac{1}{21}\) - \(\frac{1}{22}\) = \(\frac{650}{462}\)

Sn = \(\frac{325}{462}\)

O\(\)ption B
Show more

Hey, can you help me understand how you moved to 2Sn and also the first line is a bit confusing. Would really appreciate your help on this. TIA
User avatar
sj296
User avatar
Joined: 09 Jul 2014
Last visit: 19 Mar 2025
Posts: 305
Own Kudos:
397
Given Kudos: 346
Location: India
Concentration: General Management, Finance
Schools: ISB '24
Schools: ISB '24
Posts: 305
Kudos: 397
The nth term of a sequence a1, a2, a3 an is given by an = 1/(n(n+2)) 28 Nov 2023, 08:25
Kudos
Add Kudos
Bookmarks
Bookmark this Post
The general term of the series has been given as an: \(\frac{1}{(n(n+2))}\)

Now, this can be re-written as 2an = \(\frac{1}{n}\) - \(\frac{1}{(n+2)}\))....................(1)

The sum of n terms Sn = a1 + a2 .....+an......these terms can be written as per the form in equation (1)

So, if you take the 2 in the denominator to LHS, Sn becomes 2Sn.

Hope, I am able to address your queries.

abishekcmouli
sj296
Tn = \(\frac{1}{n(n+2)}\) = \(\frac{[m][fraction]1}{n} \)- \(\frac{1}{(n+2)}\)/2[/fraction][/m]

2Sn = \(\frac{1}{1.3}\) + \(\frac{1}{2.4}\)+ \(\frac{1}{3.5}\)...........................\(\frac{1}{18.20}\) + \(\frac{1}{19.21}\) + \(\frac{1}{20.22}\)

2Sn = \(\frac{1}{1}\) - \(\frac{1}{3}\) + \(\frac{1}{2}\) - \(\frac{1}{4}\) + \(\frac{1}{3}\) - \(\frac{1}{5}\) ........................\(\frac{1}{18}\) - \(\frac{1}{20}\) + \(\frac{1}{19}\)- \(\frac{1}{21}\) + \(\frac{1}{20}\) - \(\frac{1}{22}\)

2Sn = 1 + \(\frac{1}{2}\) - \(\frac{1}{21}\) - \(\frac{1}{22}\) = \(\frac{650}{462}\)

Sn = \(\frac{325}{462}\)

O\(\)ption B

Hey, can you help me understand how you moved to 2Sn and also the first line is a bit confusing. Would really appreciate your help on this. TIA
Show more
Moderators:
Bunuel
Math Expert
109763 posts
Krunaal
Tuck School Moderator
853 posts