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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?

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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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.
_________________

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

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 _________________

Thanks and Regards

Abhishek....

PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS

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