Events & Promotions
|
|
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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Nov 2007:30 AM PST
-08:30 AM PST
Learn what truly sets the UC Riverside MBA apart and how it helps in your professional growth
Nov 2211:00 AM IST
-01:00 PM IST
Do RC/MSR passages scare you? e-GMAT is conducting a masterclass to help you learn – Learn effective reading strategies Tackle difficult RC & MSR with confidence Excel in timed test environment- Nov 23
11:00 AM IST
-01:00 PM IST
Attend this free GMAT Algebra Webinar and learn how to master the most challenging Inequalities and Absolute Value problems with ease. 
Nov 2510:00 AM EST
-11:00 AM EST
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors.
15 May 2020, 09:16
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
76% (02:06) correct
24%
(02:31)
wrong
based on 198
sessions
History
Date
Time
Result
Not Attempted Yet
In a sequence of numbers \(t_1\), \(t_2\), \(t_3\), ... the difference of any two successive terms is a constant. If \(|t_8| = |t_{16}|\) and \(t_3 ≠ t_7\), what is the sum of the first 23 terms of the sequence?
A. -10
B. -4
C. 0
D. 4
E. 10
Are You Up For the Challenge: 700 Level Questions
A. -10
B. -4
C. 0
D. 4
E. 10
Are You Up For the Challenge: 700 Level Questions
15 May 2020, 11:07
Kudos
Bookmarks
nth term of the sequence = a+(n-1)d
a= first term
d= the difference of any two successive terms
\(t_3 ≠ t_7\)
\(a+2d ≠ a+6d\)
d≠0
\(|t_8| = |t_{16}|\)
\(|a+7d| = |a+15d|\)
\(Case 1-\) a+7d= a+15d
We get d=0 (not possible)
Reject this case
Case 2- a+7d= -(a+15d)
2a+22d=0 or a+11d =0
Sum of first 23 terms = Mean of 23 terms * 23
Mean of 23 terms = 12th term of the sequence = a+11d
Sum of first 23 terms = (a+11d) * 23 = 0
a= first term
d= the difference of any two successive terms
\(t_3 ≠ t_7\)
\(a+2d ≠ a+6d\)
d≠0
\(|t_8| = |t_{16}|\)
\(|a+7d| = |a+15d|\)
\(Case 1-\) a+7d= a+15d
We get d=0 (not possible)
Reject this case
Case 2- a+7d= -(a+15d)
2a+22d=0 or a+11d =0
Sum of first 23 terms = Mean of 23 terms * 23
Mean of 23 terms = 12th term of the sequence = a+11d
Sum of first 23 terms = (a+11d) * 23 = 0
Bunuel
General Discussion
15 May 2020, 18:07
Kudos
Bookmarks
In a sequence of numbers t1t1, t2t2, t3t3, ... the difference of any two successive terms is a constant. If |t8|=|t16| and t3≠t7, what is the sum of the first 23 terms of the sequence?
Tn = a + (n - 1) d, where Tn = nth term and a = first term. Here d = common difference = Tn - Tn-1.
Sn =(n/2)[2a + (n- 1)d]
given,|t8|=|t16I
or,|a+7d|=|a+15d|
case I : a+7d= a+15d
or, d=0 which is not possible because if common difference is zero each term is equal which invalidates t3≠t7 .
so, case II : a+7d= -(a+15d)
or, 2a + 22d = 0
now, the sum of the first 23 terms of the sequence
=(23/2)[2a + (23- 1)d]
=23/2[2a + 22d ]
=0
correct answer C
Tn = a + (n - 1) d, where Tn = nth term and a = first term. Here d = common difference = Tn - Tn-1.
Sn =(n/2)[2a + (n- 1)d]
given,|t8|=|t16I
or,|a+7d|=|a+15d|
case I : a+7d= a+15d
or, d=0 which is not possible because if common difference is zero each term is equal which invalidates t3≠t7 .
so, case II : a+7d= -(a+15d)
or, 2a + 22d = 0
now, the sum of the first 23 terms of the sequence
=(23/2)[2a + (23- 1)d]
=23/2[2a + 22d ]
=0
correct answer C
