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For each positive integer n, the mean of the first n terms of a sequen

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New post 27 Mar 2019, 22:05
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A
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E

Difficulty:

  65% (hard)

Question Stats:

40% (01:54) correct 60% (01:24) wrong based on 20 sessions

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Re: For each positive integer n, the mean of the first n terms of a sequen  [#permalink]

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New post 28 Mar 2019, 02:52
Bunuel wrote:
For each positive integer n, the mean of the first n terms of a sequence is n. What is the 2008th term of the sequence?

(A) 2,008
(B) 4,015
(C) 4,016
(D) 4,030,056
(E) 4,032,064


total sum/total intgers= mean
so sum = n*n = n^2
so sum of 2008th term ; ( 2008^2-2007^2) ; ( 2008+2007)*(2008-2007) ;
4015
IMO B
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For each positive integer n, the mean of the first n terms of a sequen  [#permalink]

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New post 28 Mar 2019, 03:02
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Bunuel wrote:
For each positive integer n, the mean of the first n terms of a sequence is n. What is the 2008th term of the sequence?

(A) 2,008
(B) 4,015
(C) 4,016
(D) 4,030,056
(E) 4,032,064


The series with one terms will be like {1} Average = 1
The series with two terms will be like {1, 3} Average = 2
The series with three terms will be like {1, 3, 5} Average = 3
The series with four terms will be like {1, 3, 5, 7} Average = 4
The series with five terms will be like {1, 3, 5, 7, 9} Average = 5

i.e. the resultant series is the series of CONSECUTIVE ODD integers

2008th term of series = 2008th Odd integer = 2*2008-1 = 4015

CONCEPT: nth odd integer = 2*n-1

Answer: Option B
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Re: For each positive integer n, the mean of the first n terms of a sequen  [#permalink]

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New post 13 Apr 2019, 20:47
Archit3110 wrote:
Bunuel wrote:
For each positive integer n, the mean of the first n terms of a sequence is n. What is the 2008th term of the sequence?

(A) 2,008
(B) 4,015
(C) 4,016
(D) 4,030,056
(E) 4,032,064


total sum/total intgers= mean
so sum = n*n = n^2
so sum of 2008th term ; ( 2008^2-2007^2) ; ( 2008+2007)*(2008-2007) ;
4015
IMO B


Hello Archit3110

Could you please explain to me the red part?

Kind regards!
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For each positive integer n, the mean of the first n terms of a sequen  [#permalink]

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New post 13 Apr 2019, 22:13
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1
jfranciscocuencag wrote:
Archit3110 wrote:
Bunuel wrote:
For each positive integer n, the mean of the first n terms of a sequence is n. What is the 2008th term of the sequence?

(A) 2,008
(B) 4,015
(C) 4,016
(D) 4,030,056
(E) 4,032,064


total sum/total intgers= mean
so sum = n*n = n^2
so sum of 2008th term ; ( 2008^2-2007^2) ; ( 2008+2007)*(2008-2007) ;
4015
IMO B




Could you please explain to me the red part?

Kind regards!


Given: Mean of the first n terms of the sequence = n

i.e. \(\frac{Sum-of-n-terms}{n} = n\)

i.e. Sum of n terms \(= n*n = n^2\)

But we also know that Sum of first n positive odd integers i.e. \(1+3+5+7+...........+(2n-1) = n^2\)

where \((2n-1)\) is the nth term of the series

therefore, the series talked about in this question is the series of first consecutive odd integers

now, 2008th term of the series = 2008th odd integer \(= 2*2008 - 1 = 4015\)

Or, what Archit mentioned is

2008th term = Sum of of 2008 terms - Sum of 2007 terms of the same series

But we know that sum of n terms = \(n^2\)

therefore, 2008th term = \(2008^2 - 2007^2 = (2008+2007)*(2008-2007)\)
Using property \(a^2 - b^2 = (a+b)*(a-b)\)

i.e. 2008th Term = 4015


I hope this helps!!!

jfranciscocuencag
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For each positive integer n, the mean of the first n terms of a sequen  [#permalink]

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New post 14 Apr 2019, 04:08
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jfranciscocuencag wrote:
Archit3110 wrote:
Bunuel wrote:
For each positive integer n, the mean of the first n terms of a sequence is n. What is the 2008th term of the sequence?

(A) 2,008
(B) 4,015
(C) 4,016
(D) 4,030,056
(E) 4,032,064


total sum/total intgers= mean
so sum = n*n = n^2
so sum of 2008th term ; ( 2008^2-2007^2) ; ( 2008+2007)*(2008-2007) ;
4015
IMO B




Hello Archit3110

Could you please explain to me the red part?

Kind regards!

jfranciscocuencag as rightly mentioned by GMATinsight sir in the above post ;


2008th term = Sum of of 2008 terms - Sum of 2007 terms of the same series

But we know that sum of n terms = n^2

therefore, 2008th term = 2008^2−2007^2 or say
(2008+2007)∗(2008−2007) "Using property a2−b2=(a+b)∗(a−b)a2−b2=(a+b)∗(a−b)"
2008th Term = 4015
:)
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For each positive integer n, the mean of the first n terms of a sequen   [#permalink] 14 Apr 2019, 04:08
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