Understanding Consecutive Integers and Average Calculation in Math

The average of a group of consecutive integers is equal to the average of the smallest and largest numbers.

To understand this concept, let’s start by defining what consecutive integers are

To understand this concept, let’s start by defining what consecutive integers are. Consecutive integers are a sequence of numbers that follow each other in order without any gaps. For example, 1, 2, 3, 4, 5 are consecutive integers.

Now, let’s suppose we have a group of consecutive integers. We can represent the first (smallest) integer as “n” and the last (largest) integer as “n + k”, where “k” represents the number of integers between the first and last in the sequence.

The average of the group of consecutive integers can be calculated by summing up all the integers in the sequence and dividing it by the total count of integers. Since the integers are consecutive, the sum of the integers can be given by the formula:

Sum = (n + n + k) * (k + 1) / 2

The total count of integers in the sequence is given by (k + 1) because there are “k” integers between the first and last, plus the first and last integers themselves.

Now, let’s calculate the average of the smallest and largest numbers. The average of two numbers can be calculated by summing up the two numbers and dividing it by 2. So, the average of the smallest (n) and largest (n + k) integers is:

Average = (n + n + k) / 2

According to the given statement, the average of the group of consecutive integers is equal to the average of the smallest and largest numbers. This can be represented mathematically as:

(n + n + k) * (k + 1) / 2 = (n + n + k) / 2

To solve this equation, we can simplify it by canceling out common terms. Let’s multiply both sides of the equation by 2 to eliminate the denominator:

2 * (n + n + k) * (k + 1) / 2 = 2 * (n + n + k) / 2

Simplifying further:

(n + n + k) * (k + 1) = (n + n + k)

Expanding the left side of the equation:

(n + n + k^2 + k) = n + n + k

Combining like terms:

2n + k^2 + k = 2n + k

Canceling out the 2n on both sides:

k^2 + k = k

Rearranging the equation:

k^2 + k – k = 0

k^2 = 0

This equation implies that k is equal to zero, which means there are no integers between the first and last in the sequence. Therefore, the group of consecutive integers consists of only one number.

In conclusion, the statement is true only if the group of consecutive integers contains a single number. Otherwise, if there are two or more integers in the sequence, the average of the group will not be equal to the average of the smallest and largest numbers.

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