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Consider a sequence of seven consecutive integers

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Consider a sequence of seven consecutive integers [#permalink]

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New post 09 Sep 2017, 01:16
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  25% (medium)

Question Stats:

78% (01:39) correct 22% (00:40) wrong based on 18 sessions

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Consider a sequence of seven consecutive integers. The average of the first five integers is \(n\). The average of all the seven integers is:

A. \(n\)
B. \(n+1\)
C. \(k*n\), where \(k\) is a function of \(n\)
D. \(n+\frac{2}{7}\)
E. \(n+2\)
[Reveal] Spoiler: OA

Last edited by niks18 on 09 Sep 2017, 10:58, edited 1 time in total.

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Re: Consider a sequence of seven consecutive integers [#permalink]

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New post 09 Sep 2017, 01:52
niks18 wrote:
Consider a sequence of seven consecutive integers. The average of the first five integers is \(n\). The average of all the seven integers is:

A. \(n\)
B. \(n+1\)
C. \(k*n\), where \(k\) is a function of \(n\)
D. \(n+\frac{2}{7}\)
E. \(n+2\)


In case of a set of consecutive numbers, Middle number = Average of set

Set of 5 numbers - a, a+1, a+2, a+3, a+4
Middle number = Average = a+2

but we are given that average for first 5 numbers is = n

hence a+2 = n

Set of 7 numbers - a, a+1, a+2, a+3, a+4, a+5, a+6
Middle number = Average = a+3 = (a+2) + 1 = n+1

**********************************************************************

Consecutive numbers are in A.P. and sum for such a set = number of terms * (first term + last term) / 2

If we divide the sum by number of terms then we get average. hence average = (first term + last term) / 2


Set of 5 numbers - a, a+1, a+2, a+3, a+4
Average = (first term + last term) / 2 = (a + a+4) / 2 = a+2
Average = a+2 = n

Set of 7 numbers - a, a+1, a+2, a+3, a+4, a+5, a+6
Average = (first term + last term) / 2 = (a + a+6) / 2 = a+3 = (a+2)+1
Average = n + 1
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Re: Consider a sequence of seven consecutive integers [#permalink]

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New post 10 Sep 2017, 08:52
niks18 wrote:
Consider a sequence of seven consecutive integers. The average of the first five integers is \(n\). The average of all the seven integers is:

A. \(n\)
B. \(n+1\)
C. \(k*n\), where \(k\) is a function of \(n\)
D. \(n+\frac{2}{7}\)
E. \(n+2\)


OE

Average of consecutive numbers in an ODD sequence = Middle number of the sequence
Let the seven consecutive integers be \(1,2,3,4,5,6,7\). Hence average of this sequence \(= 4\)
First five integers of the above sequence are \(1,2,3,4,5\). Average of this sequence \(= 3 = n\)
Clearly, \(4 = 3 +1 = n+1\)

Hence Option B

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Consider a sequence of seven consecutive integers [#permalink]

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New post 10 Sep 2017, 10:35
niks18 wrote:
Consider a sequence of seven consecutive integers. The average of the first five integers is \(n\). The average of all the seven integers is:

A. \(n\)
B. \(n+1\)
C. \(k*n\), where \(k\) is a function of \(n\)
D. \(n+\frac{2}{7}\)
E. \(n+2\)


average of first five integers=middle (third) term=n
average of seven integers=middle (fourth) term=n+1
B

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Re: Consider a sequence of seven consecutive integers [#permalink]

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New post 10 Sep 2017, 10:41
This is perhaps another interpretation that I thought I would suggest:

Consider a sequence of seven consecutive integers. The average of the first five integers is n.

The above can be restated as the first five integers are n-2, n-1, n, n+1, n+2. Knowing the final two numbers in the sequence are n+3 and n+4, eliminate option A.

Substitution to find the correct answer out of B to E
5 (n-2) + 6 (n-1) + 7 (n) + 8 (n+1) + 9 (n+2) + 10 (n+3) + 11 (n+4) = 56. 56/7 (total number of terms in sequence)=8. 8 is n+1, so option B

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Re: Consider a sequence of seven consecutive integers   [#permalink] 10 Sep 2017, 10:41
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