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GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2508
Given Kudos: 1053
GPA: 3.39
If p and n are positive integers, with p > n, what is the average of n
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21 May 2021, 08:20
21 May 2021, 08:20
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81% (01:40) correct
18% (01:26) wrong based on 11 sessions
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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
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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
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