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.
Sep 2807:00 PM EDT
-10:00 PM EDT
Boost your GMAT score in less than one month in a live online class + 6 months access to TTP OnDemand video courses included! Class starts Mon, Tues, Wed, Thur, Fri Sept. 28, 2026 - Oct. 16, 2027, 7:00pm-10:00pm EST
Sep 2911:00 AM EDT
-01:00 PM EDT
" Description: Top scores are possible when you enroll in a powerful GMAT course, taught live online + 6 months access to TTP OnDemand video courses included! Class starts Tues/Thur Sept. 29, 2026 - Dec. 3, 2026, 11:00am-1:00pm EST
Oct 0107:00 AM BST
-08:00 AM GMT
Welcome to the GMAT Club's Free Interview Prep Event for Top MBA Programs! Buy tickets - Virtual series that’s all set to equip you to -excel in your MBA interviews.
Oct 0301:00 PM EDT
-04:00 PM EDT
Top scores are possible when you enroll in a powerful GMAT course, taught live online + 6 months access to TTP OnDemand video courses included! Class starts Saturdays Oct. 3, 2026 - Dec. 26, 2026, 1:00pm-4:00pm EST
16 Jan 2026, 04:02
Kudos
Bookmarks
A sequence of numbers (geometric sequence) is given by the expression: g(n)=5 *(-1/2)^n.
If the sequence begins with n = 1, what are the first two terms for which |g(n) - g(n+1)| < 1/1000 ?
\(g(n) = 5*(-1/2)^n\)
g(n+1) = 5*(-1/2)^{n+1} = 5*(-1/2)^n*(-1/2) = -g(n)/2
g(n) - g(n+1) = g(n) + g(n)/2 = 3g(n)/2 = 1.5g(n)
|g(n) - g(n+1)| = 1.5|g(n)| = 7.5/2^n < 1/1000
2^n > 7.5*1000 = 7500
2^13 = 8192 > 7500
n = 13
IMO D
If the sequence begins with n = 1, what are the first two terms for which |g(n) - g(n+1)| < 1/1000 ?
\(g(n) = 5*(-1/2)^n\)
g(n+1) = 5*(-1/2)^{n+1} = 5*(-1/2)^n*(-1/2) = -g(n)/2
g(n) - g(n+1) = g(n) + g(n)/2 = 3g(n)/2 = 1.5g(n)
|g(n) - g(n+1)| = 1.5|g(n)| = 7.5/2^n < 1/1000
2^n > 7.5*1000 = 7500
2^13 = 8192 > 7500
n = 13
IMO D
10 Jun 2026, 23:27
Kudos
Bookmarks
A sequence of numbers (geometric sequence) is given by the expression: g(n)=5 *(-1/2)^n. If the sequence begins with n = 1, what are the first two terms for which |g(n) - g(n+1)| < 1/1000 ?
A. g(10), g(11)
B. g(11), g(12)
C. g(12), g(13)
D. g(13), g(14)
E. g(14), g(15)
SOLUTION: Given that g(n) = 5*(-1/2)^n, therefore
g(1) = -5/2^1
g(2) = 5/2^2
g(3) = -5/2^3
g(4) = 5/2^4
...
...
g(10) = 5/2^10
g(11) = -5/2^11
g(12) = 5/2^12
g(13) = -5/2^13
g(14) = 5/2^14
It implies that
|g(1) - g(2)| = 15/4
|g(2) - g(3)| = 15/8
|g(3) - g(4)| = 15/16
...
...
|g(10) - g(11)| = 15/2048
|g(11) - g(12)| = 15/4096
|g(12) - g(13)| = 15/8192
|g(13) - g(14)| = 15/16384 < 1/10000 (This is first set or smallest set of values for which the requirement is satisfied)
|g(14) - g(15)| = 15/32768 < 1/10000
Therefore, choice (D) is the correct one.
A. g(10), g(11)
B. g(11), g(12)
C. g(12), g(13)
D. g(13), g(14)
E. g(14), g(15)
SOLUTION: Given that g(n) = 5*(-1/2)^n, therefore
g(1) = -5/2^1
g(2) = 5/2^2
g(3) = -5/2^3
g(4) = 5/2^4
...
...
g(10) = 5/2^10
g(11) = -5/2^11
g(12) = 5/2^12
g(13) = -5/2^13
g(14) = 5/2^14
It implies that
|g(1) - g(2)| = 15/4
|g(2) - g(3)| = 15/8
|g(3) - g(4)| = 15/16
...
...
|g(10) - g(11)| = 15/2048
|g(11) - g(12)| = 15/4096
|g(12) - g(13)| = 15/8192
|g(13) - g(14)| = 15/16384 < 1/10000 (This is first set or smallest set of values for which the requirement is satisfied)
|g(14) - g(15)| = 15/32768 < 1/10000
Therefore, choice (D) is the correct one.
Moderator:
