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.
May 2008:00 AM PDT
-08:30 AM PDT
What’s in it for you- Live Profile Evaluation Chat Session with Jenifer Turtschanow, CEO, ARINGO. Come with your details prepared and ARINGO will share insights! Pre-MBA Role/Industry, YOE, Exam Score, C/GPA, ECs Post-MBA Role/ Industry & School List.
Jun 1006:00 AM PDT
-06:15 PM PDT
Register for the GMAT Club Virtual MBA Spotlight Fair – the world’s premier event for serious MBA candidates. This is your chance to hear directly from Admissions Directors at nearly every Top 30 MBA program..
12 Mar 2026, 01:41
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% (02:13) correct
24%
(03:08)
wrong
based on 58
sessions
History
Date
Time
Result
Not Attempted Yet
For the infinite sequence a1, a2, a3, ...
\(a_n = 3a_{(n−1)} − n\) for all integers n > 1.
If \(a_4 =50\), what is \(a_2 − a_1\)?
A. −4
B. 3
C. 4
D. 5
E. 7
\(a_n = 3a_{(n−1)} − n\) for all integers n > 1.
If \(a_4 =50\), what is \(a_2 − a_1\)?
A. −4
B. 3
C. 4
D. 5
E. 7
12 Mar 2026, 03:51
Kudos
Bookmarks
This is a classic Recursive Sequences — Work Backwards problem, and it's one of the GMAT Focus favorites on sequences.
The common trap: students try to work forward, calculating from a1 to a4. That doesn't work here because a1 is unknown. The moment you see a recursive formula and the ending term is given, reverse-engineer by substituting backwards.
Step 1 — Write out what you know.
The recurrence is: a_n = 3*a_(n-1) - n, for n > 1. We know a4 = 50.
Step 2 — Find a3.
a4 = 3*a3 - 4
50 = 3*a3 - 4
3*a3 = 54
a3 = 18
Step 3 — Find a2.
a3 = 3*a2 - 3
18 = 3*a2 - 3
3*a2 = 21
a2 = 7
Step 4 — Find a1.
a2 = 3*a1 - 2
7 = 3*a1 - 2
3*a1 = 9
a1 = 3
Step 5 — Compute the answer.
a2 - a1 = 7 - 3 = 4
Answer: C (4)
Takeaway: when a recursive sequence gives you a later term and asks about earlier terms, always reverse-engineer by substituting the known value backwards into the recurrence — it's almost always faster than looking for a general formula. The "-n" part of this formula changes with each step, which is exactly why forward guessing is slow and backwards substitution is clean.
The common trap: students try to work forward, calculating from a1 to a4. That doesn't work here because a1 is unknown. The moment you see a recursive formula and the ending term is given, reverse-engineer by substituting backwards.
Step 1 — Write out what you know.
The recurrence is: a_n = 3*a_(n-1) - n, for n > 1. We know a4 = 50.
Step 2 — Find a3.
a4 = 3*a3 - 4
50 = 3*a3 - 4
3*a3 = 54
a3 = 18
Step 3 — Find a2.
a3 = 3*a2 - 3
18 = 3*a2 - 3
3*a2 = 21
a2 = 7
Step 4 — Find a1.
a2 = 3*a1 - 2
7 = 3*a1 - 2
3*a1 = 9
a1 = 3
Step 5 — Compute the answer.
a2 - a1 = 7 - 3 = 4
Answer: C (4)
Takeaway: when a recursive sequence gives you a later term and asks about earlier terms, always reverse-engineer by substituting the known value backwards into the recurrence — it's almost always faster than looking for a general formula. The "-n" part of this formula changes with each step, which is exactly why forward guessing is slow and backwards substitution is clean.
12 Mar 2026, 07:31
Kudos
Bookmarks
kevincan
Let the initial term be a1.
a2 = 3a1 - 2
a3 = 3a2 - 3 = 3* (3a1-2) -3 = 9a1 - 9.
a4 = 3a3 - 4 = 3*(9a1 - 9) - 4 = 27a1 - 31.
Given that:
27a1 - 31 = 50
27a1 = 81
a1 = 3
Then, a2 - a1 = ( 3a1 -2) - a1 = 2*a1 -2 = 2*3 -2 = 4
Option C
