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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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Updated on: 09 Sep 2017, 10:58
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Difficulty:

25% (medium)

Question Stats:

79% (01:32) correct 21% (00:40) wrong based on 19 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. $$n+2$$

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Originally posted by niks18 on 09 Sep 2017, 01:16.
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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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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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
VP
Joined: 07 Dec 2014
Posts: 1018
Consider a sequence of seven consecutive integers [#permalink]

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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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Location: United States (TX)
Concentration: Strategy, General Management
GPA: 3.64
Re: Consider a sequence of seven consecutive integers [#permalink]

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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

--== Message from GMAT Club Team ==--

This is not a quality discussion. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.
Re: Consider a sequence of seven consecutive integers   [#permalink] 10 Sep 2017, 10:41
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