Last visit was: 19 Nov 2025, 08:03 It is currently 19 Nov 2025, 08:03
Home
GMAT
Messages
Tests
Schools
Events
Rewards
Chat
My Profile
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+ Level|   Sequences|      
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,389
Own Kudos:
778,260
 [15]
Given Kudos: 99,977
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: 105,389
Kudos: 778,260
 [15]
For every positive integer n ≥ 2, t1+ t2 +... + tn = 2n2 + 9n + 13. If 06 Apr 2020, 09:59
Kudos
Add Kudos
15
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
E
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:

53% (03:01) correct 47% (02:55) wrong based on 140 sessions

History

Date
Time
Result
Not Attempted Yet
For every positive integer \(n ≥ 2\), \(t_1+ t_2 +... + t_n = 2n^2 + 9n + 13\). If \(t_k=103\), then what is the value of k ?

A. 20
B. 21
C. 22
D. 23
E. 24

Are You Up For the Challenge: 700 Level Questions
ShowHide Answer
Official Answer
E
User avatar
Kinshook
User avatar
Major Poster
Joined: 03 Jun 2019
Last visit: 19 Nov 2025
Posts: 5,794
Own Kudos:
5,509
 [4]
Given Kudos: 161
Location: India
Schools: HBS '26 CBS '26 Wharton '26
GMAT 1: 690 Q50 V34
WE:Engineering (Transportation)
Products:
My Reviews
Schools: HBS '26 CBS '26 Wharton '26
GMAT 1: 690 Q50 V34
Posts: 5,794
Kudos: 5,509
 [4]
For every positive integer n ≥ 2, t1+ t2 +... + tn = 2n2 + 9n + 13. If 06 Apr 2020, 10:28
2
Kudos
Add Kudos
2
Bookmarks
Bookmark this Post
Bunuel
For every positive integer \(n ≥ 2\), \(t_1+ t_2 +... + t_n = 2n^2 + 9n + 13\). If \(t_k=103\), then what is the value of k ?

A. 20
B. 21
C. 22
D. 23
E. 24

Are You Up For the Challenge: 700 Level Questions
Show more

For every positive integer \(n ≥ 2\), \(t_1+ t_2 +... + t_n = 2n^2 + 9n + 13\). If \(t_k=103\), then what is the value of k ?

\(t_1+ t_2 +... + t_n = 2n^2 + 9n + 13\)
\(t_1+ t_2 +... + t_{n-1} = 2(n-1)^2 + 9(n-1) + 13\)

\(t_n = 2n^2 + 9n + 13 - [2(n-1)^2 + 9(n-1) + 13] = 2(1)(2n-1) + 9 = 4n -2 + 9 = 4n +7\)

Since \(t_k = 103 = 4n + 7 \)
n = 24

IMO E
User avatar
Ahmed9955
User avatar
Joined: 18 Feb 2019
Last visit: 02 Dec 2023
Posts: 83
Own Kudos:
24
Given Kudos: 326
Location: India
Schools: Booth '24 Fuqua '24 ISB '23 NTU NUS Ross '24 Rotman '24 Yale '24
GMAT 1: 570 Q46 V21
Schools: Booth '24 Fuqua '24 ISB '23 NTU NUS Ross '24 Rotman '24 Yale '24
GMAT 1: 570 Q46 V21
Posts: 83
Kudos: 24
For every positive integer n ≥ 2, t1+ t2 +... + tn = 2n2 + 9n + 13. If 08 Apr 2020, 01:22
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Kinshook
Bunuel
For every positive integer \(n ≥ 2\), \(t_1+ t_2 +... + t_n = 2n^2 + 9n + 13\). If \(t_k=103\), then what is the value of k ?

A. 20
B. 21
C. 22
D. 23
E. 24

Are You Up For the Challenge: 700 Level Questions

For every positive integer \(n ≥ 2\), \(t_1+ t_2 +... + t_n = 2n^2 + 9n + 13\). If \(t_k=103\), then what is the value of k ?

\(t_1+ t_2 +... + t_n = 2n^2 + 9n + 13\)
\(t_1+ t_2 +... + t_{n-1} = 2(n-1)^2 + 9(n-1) + 13\)

\(t_n = 2n^2 + 9n + 13 - [2(n-1)^2 + 9(n-1) + 13] = 2(1)(2n-1) + 9 = 4n -2 + 9 = 4n +7\)

Since \(t_k = 103 = 4n + 7 \)
n = 24

IMO E
Show more


hi.
can you please explain highlighted red part.
how is tk= tn-t(n-1)?
User avatar
gurmukh
User avatar
Joined: 18 Dec 2017
Last visit: 24 Jan 2025
Posts: 259
Own Kudos:
260
Given Kudos: 20
Posts: 259
Kudos: 260
For every positive integer n ≥ 2, t1+ t2 +... + tn = 2n2 + 9n + 13. If 08 Apr 2020, 02:39
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Ahmed9955
Kinshook
Bunuel
For every positive integer \(n ≥ 2\), \(t_1+ t_2 +... + t_n = 2n^2 + 9n + 13\). If \(t_k=103\), then what is the value of k ?

A. 20
B. 21
C. 22
D. 23
E. 24

Are You Up For the Challenge: 700 Level Questions

For every positive integer \(n ≥ 2\), \(t_1+ t_2 +... + t_n = 2n^2 + 9n + 13\). If \(t_k=103\), then what is the value of k ?

\(t_1+ t_2 +... + t_n = 2n^2 + 9n + 13\)
\(t_1+ t_2 +... + t_{n-1} = 2(n-1)^2 + 9(n-1) + 13\)

\(t_n = 2n^2 + 9n + 13 - [2(n-1)^2 + 9(n-1) + 13] = 2(1)(2n-1) + 9 = 4n -2 + 9 = 4n +7\)

Since \(t_k = 103 = 4n + 7 \)
n = 24

IMO E
Show more


hi.
can you please explain highlighted red part.
how is tk= tn-t(n-1)?

It is just like from sum of 24 terms if you subtract sum of 23 terms in a series so you get 24 th term
Now 24term is given to us as 103 and we also know its value in terms of n so we have equated to get 24.

Posted from my mobile device
User avatar
Ahmed9955
User avatar
Joined: 18 Feb 2019
Last visit: 02 Dec 2023
Posts: 83
Own Kudos:
24
Given Kudos: 326
Location: India
Schools: Booth '24 Fuqua '24 ISB '23 NTU NUS Ross '24 Rotman '24 Yale '24
GMAT 1: 570 Q46 V21
Schools: Booth '24 Fuqua '24 ISB '23 NTU NUS Ross '24 Rotman '24 Yale '24
GMAT 1: 570 Q46 V21
Posts: 83
Kudos: 24
For every positive integer n ≥ 2, t1+ t2 +... + tn = 2n2 + 9n + 13. If 08 Apr 2020, 02:56
Kudos
Add Kudos
Bookmarks
Bookmark this Post
hi.
can you please explain highlighted red part.
how is tk= tn-t(n-1)?

It is just like from sum of 24 terms if you subtract sum of 23 terms in a series so you get 24 th term
Now 24term is given to us as 103 and we also know its value in terms of n so we have equated to get 24.

Posted from my mobile device[/quote]

hey Gurmukh ,
how do we know 24th term is 103?
where is it given?
User avatar
gurmukh
User avatar
Joined: 18 Dec 2017
Last visit: 24 Jan 2025
Posts: 259
Own Kudos:
260
Given Kudos: 20
Posts: 259
Kudos: 260
For every positive integer n ≥ 2, t1+ t2 +... + tn = 2n2 + 9n + 13. If 08 Apr 2020, 03:12
Kudos
Add Kudos
Bookmarks
Bookmark this Post
No, we don't know Tk is 103 and I have taken 23 just for explaining sake. The point is from the sum of first tk terms if we subtract t(k-1) terms then we get tk th term.

Posted from my mobile device
User avatar
ScottTargetTestPrep
User avatar
Target Test Prep Representative
Joined: 14 Oct 2015
Last visit: 18 Nov 2025
Posts: 21,713
Own Kudos:
26,995
 [2]
Given Kudos: 300
Status:Founder & CEO
Affiliations: Target Test Prep
Location: United States (CA)
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 21,713
Kudos: 26,995
 [2]
For every positive integer n ≥ 2, t1+ t2 +... + tn = 2n2 + 9n + 13. If 09 Apr 2020, 04:36
1
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Bunuel
For every positive integer \(n ≥ 2\), \(t_1+ t_2 +... + t_n = 2n^2 + 9n + 13\). If \(t_k=103\), then what is the value of k ?

A. 20
B. 21
C. 22
D. 23
E. 24

Are You Up For the Challenge: 700 Level Questions
Show more

We see that:

t(1) + t(2) + … + t(k - 1) + t(k) = 2k^2 + 9k + 13

and

t(1) + t(2) + … + t(k - 1) = 2(k - 1)^2 + 9(k - 1) + 13

If we subtract the second equation from the first, we have:

t(k) = 2k^2 + 9k - [2(k - 1)^2 + 9(k - 1)]

Since we are given that t(k) = 103, we have:

103 = 2k^2 + 9k - [2k^2 - 4k + 2 + 9k - 9]

103 = -[-4k + 2 - 9]

103 = 4k + 7

96 = 4k

24 = k

Answer: E
User avatar
exc4libur
User avatar
Joined: 24 Nov 2016
Last visit: 22 Mar 2022
Posts: 1,684
Own Kudos:
1,447
Given Kudos: 607
Location: United States
Posts: 1,684
Kudos: 1,447
For every positive integer n ≥ 2, t1+ t2 +... + tn = 2n2 + 9n + 13. If 11 May 2020, 12:44
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
For every positive integer \(n ≥ 2\), \(t_1+ t_2 +... + t_n = 2n^2 + 9n + 13\). If \(t_k=103\), then what is the value of k ?

A. 20
B. 21
C. 22
D. 23
E. 24

Are You Up For the Challenge: 700 Level Questions
Show more

t1+t2=2(n^2)+9n+13=2*4+9*4+13=39
t1+t2+t3=2(3^2)+9(3)+13=18+27+13=58
t1+t2+t3+t4=2(4^2)+9(4)+13=81
t1+t2+t3+t4+t5=2(5^2)+9(5)+13=108
t3=58-39=19
t4=81-58=23=t3+4
t5=108-81=27=t4+4=t3+8
tk=t3+4(k-3)
tk: 103=19+4(k-3), 4k=103-19+12, k=96/4=24

ans (E)
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 38,588
Own Kudos:
1,079
Posts: 38,588
Kudos: 1,079
For every positive integer n ≥ 2, t1+ t2 +... + tn = 2n2 + 9n + 13. If 25 Sep 2025, 04:27
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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.
Moderators:
Bunuel
Math Expert
105389 posts
Krunaal
Tuck School Moderator
805 posts