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# There are 9 consecutive integers in a certain sequence. If the average

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Joined: 03 Feb 2017
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There are 9 consecutive integers in a certain sequence. If the average [#permalink]

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16 Jun 2017, 05:23
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There are 9 consecutive integers in a certain sequence. If the average of the first seven integers in the sequence is n, what is the average of all 9 integers in the sequence?

A) n
B) n-1
C) n+1
D) 2n
E) n(n+1)
[Reveal] Spoiler: OA
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Joined: 26 Feb 2016
Posts: 2442
Location: India
GPA: 3.12
Re: There are 9 consecutive integers in a certain sequence. If the average [#permalink]

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16 Jun 2017, 06:09
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KUDOS
Consider the 9 consecutive number 1,2,3,4,5,6,7,8,9

The average of the first seven integers is 4(Middle term, since elements are in AP)
The average of the entire lot will be 5(which is n+1, provided the average of first 7 numbers is n)

Hence, Option C is the answer option.
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Re: There are 9 consecutive integers in a certain sequence. If the average [#permalink]

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16 Jun 2017, 06:47
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danlew wrote:
There are 9 consecutive integers in a certain sequence. If the average of the first seven integers in the sequence is n, what is the average of all 9 integers in the sequence?

A) n
B) n-1
C) n+1
D) 2n
E) n(n+1)

USEFUL RULE: In a set of EQUALLY SPACED numbers, the mean of the set equals the median of the set

Let O, O, O, O, O, O, O, O, O represent the 9 consecutive integers listed in ascending order.

The average (aka MEAN) of the first seven integers in the sequence is n
Here are the first seven integers: O, O, O, O, O, O, O
Since consecutive integers are equally spaced, the mean = the median
The median of these 7 numbers is the middlemost number.
So, the middlemost value is also the mean.
We get: O, O, O, n, O, O, O
So, our NINE numbers now look like this: O, O, O, n, O, O, O, O, O

What is the average of all 9 integers in the sequence?
Using the same logic, we can see that the mean of the 9 numbers is the middlemost number.
So, replace the middlemost number with a "?" : O, O, O, n, ?, O, O, O, O

We can see that ? (the mean of all 9 integers) is 1 GREATER THAN n (the mean of the first 7 integers)
In other words, the mean of all 9 integers = n+1

[Reveal] Spoiler:
C

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Re: There are 9 consecutive integers in a certain sequence. If the average [#permalink]

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16 Jun 2017, 09:03
danlew wrote:
There are 9 consecutive integers in a certain sequence. If the average of the first seven integers in the sequence is n, what is the average of all 9 integers in the sequence?

A) n
B) n-1
C) n+1
D) 2n
E) n(n+1)

Plug in and check...

Let the sequence of consecutive integers be 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9

Sum of first 7 numbers is 28
Average of first 7 numbers is $$\frac{28}{7} = 4$$

Sum of first 9 numbers is 45
Average of first 9 numbers is $$\frac{45}{9} = 5$$

Hence, the answer will be (C) n+1
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Director
Joined: 07 Dec 2014
Posts: 965
There are 9 consecutive integers in a certain sequence. If the average [#permalink]

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16 Jun 2017, 14:04
danlew wrote:
There are 9 consecutive integers in a certain sequence. If the average of the first seven integers in the sequence is n, what is the average of all 9 integers in the sequence?

A) n
B) n-1
C) n+1
D) 2n
E) n(n+1)

if a=average of all 9 consecutive integers,
a will be the 5th term
n will be the 4th term
a=n+(5-4)=n+1
C
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Joined: 12 Oct 2017
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There are 9 consecutive integers in a certain sequence. If the average [#permalink]

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16 Dec 2017, 07:55
Since we know n is average for 7, for 9 consecutive numbers it would be n+ (n+1) +(n+2) / 3 = 3n+3/3. which is n+1.
Ans C.
There are 9 consecutive integers in a certain sequence. If the average   [#permalink] 16 Dec 2017, 07:55
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