Last visit was: 01 May 2026, 11:23 It is currently 01 May 2026, 11:23
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
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 01 May 2026
Posts: 109,997
Own Kudos:
812,324
 [41]
Given Kudos: 105,974
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: 109,997
Kudos: 812,324
 [41]
In the sequence above, each term after the first term is equal to the 13 May 2015, 04:47
4
Kudos
Add Kudos
37
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

25% (medium)

Question Stats:

84% (02:32) correct 16% (03:08) wrong based on 885 sessions

History

Date
Time
Result
Not Attempted Yet
\(t_1\), \(t_2\), \(t_3\), ..., \(t_n\), ...
In the sequence above, each term after the first term is equal to the preceding term plus the constant k. If \(t_1+t_3+t_5+t_7=32\), what is the value of \(t_2+t_4+t_6\) ?

A. 8
B. 12
C. 24
D. 32
E. 72

Experience GMAT Club Test Questions
Yes, you've landed on a GMAT Club Tests question
Craving more? Unlock our full suite of GMAT Club Tests here
Want to experience more? Get a taste of our tests with our free trial today
Rise to the challenge with GMAT Club Tests. Happy practicing!
GMAT Club Tests Check Our Products Try Free Tests
ShowHide Answer
Official Answer
C
Most Helpful Reply
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 01 May 2026
Posts: 109,997
Own Kudos:
812,324
 [6]
Given Kudos: 105,974
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: 109,997
Kudos: 812,324
 [6]
In the sequence above, each term after the first term is equal to the 18 May 2015, 06:29
4
Kudos
Add Kudos
2
Bookmarks
Bookmark this Post
Bunuel
\(t_1\), \(t_2\), \(t_3\), ..., \(t_n\), ...
In the sequence above, each term after the first term is equal to the preceding term plus the constant k. If \(t_1+t_3+t_5+t_7=32\), what is the value of \(t_2+t_4+t_6\) ?

A. 8
B. 12
C. 24
D. 32
E. 72

Kudos for a correct solution.
Show more

OFFICIAL SOLUTION:

According to the stem:
\(t_2=t_1+k\);
\(t_3=t_2+k=t_1+2k\);
\(t_4=t_3+k=t_1+3k\);
...
\(t_n=t_1+(n-1)k\);

Since \(t_1+t_3+t_5+t_7=32\), then \(t_1+(t_1+2k)+(t_1+4k)+(t_1+6k)=32\). So, \(t_1+3k=8\)

We need to find the of \(t_2+t_4+t_6=(t_1+k)+(t_1+3k)+(t_1+5k)=3(t_1+3k)\). From above we know that \(t_1+3k=8\), thus \(3(t_1+3k)=3*8=24\).

Answer: C.
User avatar
KarishmaB
User avatar
Joined: 16 Oct 2010
Last visit: 29 Apr 2026
Posts: 16,448
Own Kudos:
79,467
 [5]
Given Kudos: 485
Location: Pune, India
Products:
GMAT Club Tests
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,448
Kudos: 79,467
 [5]
In the sequence above, each term after the first term is equal to the 03 Jul 2016, 20:27
3
Kudos
Add Kudos
2
Bookmarks
Bookmark this Post
Bunuel
\(t_1\), \(t_2\), \(t_3\), ..., \(t_n\), ...
In the sequence above, each term after the first term is equal to the preceding term plus the constant k. If \(t_1+t_3+t_5+t_7=32\), what is the value of \(t_2+t_4+t_6\) ?

A. 8
B. 12
C. 24
D. 32
E. 72

Kudos for a correct solution.
Show more


Another method is to note that the average of the two sequences would be the same.

The average of \(t_1\), \(t_3\), \(t_5\) and \(t_7\) will be the middle of \(t_3\) and \(t_5\) which is \(t_4\). (All terms in an AP are equidistant from the term before it and the term after it.)
The average of \(t_2\), \(t_4\), \(t_6\) is obviously \(t_4\).

\(t_1+t_3+t_5+t_7=4 * t_4 = 32\)

\(t_4 = 8\)

Then \(t_2+t_4+t_6 = 3 * t_4 = 24\)

Answer (C)
General Discussion
User avatar
peachfuzz
User avatar
Joined: 28 Feb 2014
Last visit: 27 Jan 2018
Posts: 266
Own Kudos:
369
 [3]
Given Kudos: 132
Location: United States
Concentration: Strategy, General Management
Products:
My Reviews
Posts: 266
Kudos: 369
 [3]
In the sequence above, each term after the first term is equal to the 13 May 2015, 07:03
2
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
x+x+2k+x+4k+x+6k=32
4x+12k=32
x+3k=8

We are looking for the value of:
x+k+x+3k+x+5k=?
3x+9k=?

3(x+3k=8)
3x+9k=24

Answer: C
avatar
htridn
avatar
Joined: 22 Mar 2015
Last visit: 14 Apr 2017
Posts: 3
Own Kudos:
5
 [2]
Given Kudos: 2
Posts: 3
Kudos: 5
 [2]
In the sequence above, each term after the first term is equal to the 13 May 2015, 20:11
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
t1 + t3 +t5 +t7 = t1 + t1 +2k +t1 +4k +t1 +6k= 4t + 12k = 32
t = 2 ==> k =2 ==> t2+t4 +t6 = 4 +8 +12 = 24
=> C
User avatar
D3N0
User avatar
Joined: 21 Jan 2015
Last visit: 28 Apr 2026
Posts: 585
Own Kudos:
608
 [2]
Given Kudos: 132
Location: India
Concentration: Operations, Technology
Schools: Smeal'21 (WD) Arizona State'21 (A$) Simon '21 (WD)
GMAT 1: 620 Q48 V28
GMAT 2: 690 Q49 V35
WE:Operations (Retail: E-commerce)
Products:
My Reviews GMAT Club Tests
Schools: Smeal'21 (WD) Arizona State'21 (A$) Simon '21 (WD)
GMAT 2: 690 Q49 V35
Posts: 585
Kudos: 608
 [2]
In the sequence above, each term after the first term is equal to the 13 May 2015, 21:59
1
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Bunuel
\(t_1\), \(t_2\), \(t_3\), ..., \(t_n\), ...
In the sequence above, each term after the first term is equal to the preceding term plus the constant k. If \(t_1+t_3+t_5+t_7=32\), what is the value of \(t_2+t_4+t_6\) ?

A. 8
B. 12
C. 24
D. 32
E. 72
Show more

Ans: C
Solution: according to given condition
t2=t1+k
t3=t2+k = t1+2k
same way t4=t1+3k
t5=t1+4k
t6=t1+5k
t7=t1+6k

now by putting the values in given equation
t1+t3+t5+t7=32
t1+3k=8

required value
t2+t4+t6 = 3(t1+3k)= 3*8= 24

Ans: C
User avatar
iqbalfiery
User avatar
Joined: 21 Feb 2016
Last visit: 03 Jan 2021
Posts: 9
Own Kudos:
2
 [1]
Given Kudos: 107
Location: United States (MA)
Posts: 9
Kudos: 2
 [1]
In the sequence above, each term after the first term is equal to the 03 Jul 2016, 04:58
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
\(t_1\), \(t_2\), \(t_3\), ..., \(t_n\), ...
In the sequence above, each term after the first term is equal to the preceding term plus the constant k. If \(t_1+t_3+t_5+t_7=32\), what is the value of \(t_2+t_4+t_6\) ?

A. 8
B. 12
C. 24
D. 32
E. 72

Kudos for a correct solution.
Show more

According to the question, t2=t1+k, t3=t2+k=t1+k+k So, t3=t1+2k, t4=t1+3k, t5=t1+4k.. That's the pattern.
So, t1+t3+t5+t7
=t1+t1+2k+t1+4k+t1+6k
=4t+12k which is 32 given
At this point, u can take any value of t and k that satisfies the equation 4t+12k=32. (The (t,k) pairs could be (2,2) or (5,1).
Now, we are asked to find t2+t4+t6, which can be written as 3t+k+3k+5k, which in turn is equal to 3t+9k.
Taking either pair of values of (t,k), we get 3t+9k=6+18=24.
So answer is B

Alternatively, you could see that as t1+t3+t5+t7=32, and as 1+3+5+7=16, so the total, to be 32, each term has to be double, i.e. 2+6+10+14=32. Now these terms are not consecutive meaning there is one term between any two consecutive of these terms. So, t2+t4+t6 would be 4+8+12, which is 24.
Hope it helps.
User avatar
gracie
User avatar
Joined: 07 Dec 2014
Last visit: 11 Oct 2020
Posts: 1,028
Own Kudos:
2,027
Given Kudos: 27
Posts: 1,028
Kudos: 2,027
In the sequence above, each term after the first term is equal to the Updated on: 07 Jul 2018, 20:58
Originally posted by gracie on 03 Jul 2016, 19:58.
Last edited by gracie on 07 Jul 2018, 20:58, edited 1 time in total.
Kudos
Add Kudos
Bookmarks
Bookmark this Post
t1t1, t2t2, t3t3, ..., tntn, ...
In the sequence above, each term after the first term is equal to the preceding term plus the constant k. If t1+t3+t5+t7=32t1+t3+t5+t7=32, what is the value of t2+t4+t6t2+t4+t6 ?

A. 8
B. 12
C. 24
D. 32
E. 72


t1+t3+t5+t7=32
➡4*t1+12k=32
➡t1+3k=8
t1=2
k=2
t2+t4+t6=4+8+12=24
C
avatar
sk1007
avatar
Joined: 26 Jul 2017
Last visit: 21 Oct 2018
Posts: 5
Own Kudos:
2
 [1]
Posts: 5
Kudos: 2
 [1]
In the sequence above, each term after the first term is equal to the 06 Jul 2018, 22:54
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
Bunuel
\(t_1\), \(t_2\), \(t_3\), ..., \(t_n\), ...
In the sequence above, each term after the first term is equal to the preceding term plus the constant k. If \(t_1+t_3+t_5+t_7=32\), what is the value of \(t_2+t_4+t_6\) ?

A. 8
B. 12
C. 24
D. 32
E. 72

Kudos for a correct solution.

OFFICIAL SOLUTION:

According to the stem:
\(t_2=t_1+k\);
\(t_3=t_2+k=t_1+2k\);
\(t_4=t_3+k=t_1+3k\);
...
\(t_n=t_1+(n-1)k\);

Since \(t_1+t_3+t_5+t_7=32\), then \(t_1+(t_1+2k)+(t_1+4k)+(t_1+6k)=32\). So, \(t_1+3k=8\)

We need to find the of \(t_2+t_4+t_6=(t_1+k)+(t_1+3k)+(t_1+5k)=3(t_1+3k)\). From above we know that \(t_1+3k=8\), thus \(3(t_1+3k)=3*8=24\).

Answer: C.
Show more


What about the box symbol that is in front of t3, t5 and t4? How are we saying that t3 = box t3?
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 01 May 2026
Posts: 109,997
Own Kudos:
812,324
Given Kudos: 105,974
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: 109,997
Kudos: 812,324
In the sequence above, each term after the first term is equal to the 07 Jul 2018, 00:29
Kudos
Add Kudos
Bookmarks
Bookmark this Post
sk1007
Bunuel
Bunuel
\(t_1\), \(t_2\), \(t_3\), ..., \(t_n\), ...
In the sequence above, each term after the first term is equal to the preceding term plus the constant k. If \(t_1+t_3+t_5+t_7=32\), what is the value of \(t_2+t_4+t_6\) ?

A. 8
B. 12
C. 24
D. 32
E. 72

Kudos for a correct solution.

OFFICIAL SOLUTION:

According to the stem:
\(t_2=t_1+k\);
\(t_3=t_2+k=t_1+2k\);
\(t_4=t_3+k=t_1+3k\);
...
\(t_n=t_1+(n-1)k\);

Since \(t_1+t_3+t_5+t_7=32\), then \(t_1+(t_1+2k)+(t_1+4k)+(t_1+6k)=32\). So, \(t_1+3k=8\)

We need to find the of \(t_2+t_4+t_6=(t_1+k)+(t_1+3k)+(t_1+5k)=3(t_1+3k)\). From above we know that \(t_1+3k=8\), thus \(3(t_1+3k)=3*8=24\).

Answer: C.


What about the box symbol that is in front of t3, t5 and t4? How are we saying that t3 = box t3?
Show more

That was some bug. Edited. Thank you.
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 39,013
Own Kudos:
1,122
Posts: 39,013
Kudos: 1,122
In the sequence above, each term after the first term is equal to the 23 May 2025, 04:50
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Automated notice from GMAT Club BumpBot:

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

This post was generated automatically.
Moderators:
Bunuel
Math Expert
109997 posts
Krunaal
Tuck School Moderator
852 posts