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, 10:05
Kudos
Bookmarks
D
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:
42% (02:45) correct
58%
(03:16)
wrong
based on 52
sessions
History
Date
Time
Result
Not Attempted Yet
In a sequence of numbers in which each term after the first is found by multiplying the previous one by 1/n, where n is a positive integer, the second term is 1,000. If \(P_n\) is the product of the first n terms of the sequence, \(P_6 > P_5\) and \(P_6 > P_7\), what is the sum of all possible values of n?
A. 4
B. 5
C. 7
D. 9
E. 13
Are You Up For the Challenge: 700 Level Questions
A. 4
B. 5
C. 7
D. 9
E. 13
Are You Up For the Challenge: 700 Level Questions
answer D
n^4<1000 and n^5>1000
only numbers 4 and 5 satisfy the given condition, so sum of all number is nine.
n^4<1000 and n^5>1000
only numbers 4 and 5 satisfy the given condition, so sum of all number is nine.
15 May 2020, 10:33
Kudos
Bookmarks
Suppose first term = a
\(\frac{a}{n} =1000\)
a=1000n
\(P_6>P_5\)
\(\frac{a^6}{n^{15}} > \frac{a^5}{n^{10}}\)
Since a,n>0
\(a>n^5\).......(1)
\(P_6>P_7\)
\(\frac{a^6}{n^{15}} > \frac{a^7}{n^{21}}\)
\(a<n^6\).........(2)
\(n^5<a<n^6\) [From (1) and (2)]
\(n^5<1000n<n^6\)
\(n^4<1000<n^5\)
n can be 4 or 5
Sum of all possible values of n = 4+5=9
\(\frac{a}{n} =1000\)
a=1000n
\(P_6>P_5\)
\(\frac{a^6}{n^{15}} > \frac{a^5}{n^{10}}\)
Since a,n>0
\(a>n^5\).......(1)
\(P_6>P_7\)
\(\frac{a^6}{n^{15}} > \frac{a^7}{n^{21}}\)
\(a<n^6\).........(2)
\(n^5<a<n^6\) [From (1) and (2)]
\(n^5<1000n<n^6\)
\(n^4<1000<n^5\)
n can be 4 or 5
Sum of all possible values of n = 4+5=9
Bunuel
