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There are 9 consecutive integers in a certain sequence. If the average
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12 May 2021, 08:54
12 May 2021, 08:54
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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)
A) n
B) n-1
C) n+1
D) 2n
E) n(n+1)
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Re: There are 9 consecutive integers in a certain sequence. If the average
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12 May 2021, 09:03
12 May 2021, 09:03
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Carcass 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)
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
Answer: C
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There are 9 consecutive integers in a certain sequence. If the average
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06 Jul 2021, 20:05
06 Jul 2021, 20:05
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Theory: In case of an Arithmetic Sequence (Equally spaced numbers), Mean of n terms is same as mean of first term and last term
Average of First Seven Integers
= Average of \(1^{st}\) and \(7^{th}\) integer [ Ex 1,2,3,4,5,6,7 -> Average is \(\frac{(1+7)}{2}\) = 4 ]
= \(4^{th}\) Integer = n (given)
Average of First Nine Integers
= Average of \(1^{st}\) and \(9^{th}\) integer
= \(5^{th}\) Integer
Since they are consecutive integers, so \(5^{th}\) = \(4^{th}\) + 1 = n + 1
So, Average of First Nine Integers = n + 1
So, answer will be C
Hope it helps!
To learn more about Sequences (Arithmetic and Geometric Sequence) watch the following video
Average of First Seven Integers
= Average of \(1^{st}\) and \(7^{th}\) integer [ Ex 1,2,3,4,5,6,7 -> Average is \(\frac{(1+7)}{2}\) = 4 ]
= \(4^{th}\) Integer = n (given)
Average of First Nine Integers
= Average of \(1^{st}\) and \(9^{th}\) integer
= \(5^{th}\) Integer
Since they are consecutive integers, so \(5^{th}\) = \(4^{th}\) + 1 = n + 1
So, Average of First Nine Integers = n + 1
So, answer will be C
Hope it helps!
To learn more about Sequences (Arithmetic and Geometric Sequence) watch the following video
