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Que: If set N consists of odd numbers of consecutive integers, starting with 1, what is the difference between...

by Max@Math Revolution » Sat Dec 26, 2020 11:12 pm

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Que: If set N consists of odd numbers of consecutive integers, starting with 1, what is the difference between the average of the odd integers and the average of the even integers in set N?

(A) −1
(B) \(\frac{1}{2}\)
(C) 0
(D) 1
(E) 2
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Re: Que: If set N consists of odd numbers of consecutive integers, starting with 1, what is the difference between...

by Max@Math Revolution » Wed Dec 30, 2020 11:52 pm
Solution: Since the set starts with an odd number (1) and has an odd number of integers, the set would end with an odd number, too.

Let’s see the set. Set N: {1, 2, 3, 4, 5, ..., (2n + 1)}, where n is a positive integer.

=> Number of odd terms is one more than the number of even terms.

Thus, the number of odd terms = (n + 1) and we derive a new set from set N, that is {1,3,…, 2n+1}.

Then we get the average of the odd integers = \(\frac{\left[1\ +\ 3\ +\ ......\ +\ \left(2n\ +\ 1\right)\right]}{\left(n\ +\ 1\right)}\) and since \(1\ +\ 3\ +\ ......\ +\ \left(2n\ +\ 1\right)\ =\ \left(n\ +\ 1\right)^2\), we get \(\frac{\left[1\ +\ 3\ +\ ......\ +\ \left(2n\ +\ 1\right)\right]}{\left(n\ +\ 1\right)}=\frac{\left(n\ +1\right)^2}{\left(n\ +\ 1\right)}\ =\ n\ +\ 1\)

The number of even terms = n and we derive a new set from set N, that is {2,4,…, 2n}.

Then we get the average of the even integers = \(\frac{\left(2\ +\ 4\ +\ ....\ +\ 2n\right)}{n}=\frac{2\cdot\left(1\ +\ 2\ +\ ....\ +\ n\right)}{n}\) and since \(1\ +\ 2\ +\ ........\ +\ n\ =\ \frac{n\left(n\ +\ 1\right)}{2}\), we get \(\frac{\left(2\ +\ 4\ +\ .......\ +\ 2n\right)}{n}=\frac{2\cdot\left(1\ +\ 2\ +\ .......\ +\ n\right)}{n}=\frac{n\left(n\ +\ 1\right)}{n}\ =\ n\ +1\)

Therefore, we get the difference of the average of the odd integers and the average of the even integers in set N=(n + 1) - (n + 1) = 0

Say there are only three terms in the set

S = {1, 2, 3}

=> X = \(\frac{\left(1\ +\ 3\right)}{2}=\frac{4}{2}=2\)

=> Y = 2

=> X – Y = 0

Again, say there are five terms in the set

S = {1, 2, 3, 4, 5}

=> X = \(\frac{\left(1\ +\ 3\ +\ 5\right)}{3}=\frac{9}{3}=3\)

=> Y = \(\frac{\left(2\ +\ 4\right)}{2}=\frac{6}{2}=3\)

=> X – Y = 0

Therefore, C is the correct answer.

Answer C
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Re: Que: If set N consists of odd numbers of consecutive integers, starting with 1, what is the difference between...

by Scott@TargetTestPrep » Mon Apr 26, 2021 6:34 am
Max@Math Revolution wrote: ↑
Sat Dec 26, 2020 11:12 pm
 Que: If set N consists of odd numbers of consecutive integers, starting with 1, what is the difference between the average of the odd integers and the average of the even integers in set N?

(A) −1
(B) \(\frac{1}{2}\)
(C) 0
(D) 1
(E) 2
Solution:

We can let set N be {1, 2, 3}. So the average of the odd integers is (1 + 3)/2 = 2, and the average of the even integers is 2. We see that the difference between the two averages is 0. Although we can say the correct choice must be B, let’s make sure it is the case by using another example.

If we let set N be {1, 2, 3, 4, 5}, the average of the odd integers is (1 + 3 + 5)/3 = 3 and the average of the even integers is (2 + 4)/2 = 3. Again we see that the difference between the two averages is 0.

Answer: C

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