GRE Prep 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.
GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2506
Given Kudos: 1053
GPA: 3.39
If p and n are positive integers, with p > n, what is the average of n
[#permalink]
21 May 2021, 08:20
21 May 2021, 08:20
Expert Reply
4
Bookmarks
00:00
Question Stats:
81% (01:40) correct
18% (01:26) wrong based on 11 sessions
Hide Show timer Statistics
If \(p\) and \(n\) are positive integers, with \(p > n,\) what is the average of \(n\) consecutive multiples of \(p\) (starting with \(p\)), less the average of \(p\) consecutive multiples of \(n\) (starting with \(n\))?
A. \(\frac{(p + n)}{2}\)
B. \(\frac{(p – n)}{2}\)
C. \(p + n\)
D. \(p – n\)
E. \(p^2 – n^2\)
A. \(\frac{(p + n)}{2}\)
B. \(\frac{(p – n)}{2}\)
C. \(p + n\)
D. \(p – n\)
E. \(p^2 – n^2\)
Re: If p and n are positive integers, with p > n, what is the average of n
[#permalink]
22 May 2021, 04:38
22 May 2021, 04:38
1
Sum of n consecutive multiples of p = p*1+p*2+p*3+....p*n
=p(1+2+3+4....+n)
=p*n*(n+1)/2
Average of n consecutive multiples of p = p*n*(n+1)/2n
=p*(n+1)/2
Sum of p consecutive multiples of n = n*1+n*2+n*3+....n*p
=n(1+2+3+4....+p)
=n*p*(p+1)/2
Average of p consecutive multiples of n = n*p*(p+1)/2p
=n*(p+1)/2
Difference = (p-n)/2
=p(1+2+3+4....+n)
=p*n*(n+1)/2
Average of n consecutive multiples of p = p*n*(n+1)/2n
=p*(n+1)/2
Sum of p consecutive multiples of n = n*1+n*2+n*3+....n*p
=n(1+2+3+4....+p)
=n*p*(p+1)/2
Average of p consecutive multiples of n = n*p*(p+1)/2p
=n*(p+1)/2
Difference = (p-n)/2
gmatclubot
Re: If p and n are positive integers, with p > n, what is the average of n [#permalink]
22 May 2021, 04:38
Moderators:
