Understanding the Concept of Average in a Group of Consecutive Integers

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 consider a group of consecutive integers

To understand this concept, let’s consider a group of consecutive integers. Consecutive integers are simply numbers that follow each other in sequence without any gaps or jumps. For example, 1, 2, 3, 4, 5 are consecutive integers.

Now, let’s assume we have a group of consecutive integers. The average of these integers means finding the sum of all the integers and dividing it by the number of integers in the group.

Let’s say the group of consecutive integers starts from ‘a’ and ends at ‘b’. So, the total number of integers in this group would be (b – a + 1) since we need to include both ‘a’ and ‘b’ in the group.

The average of these consecutive integers can be represented as:
Average = (Sum of the Integers)/(Number of Integers) = (a + a+1 + a+2 + … + b)/(b – a + 1)

Now, 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 as:
Average = (Smallest Number + Largest Number)/2

We can set up an equation equating these two average expressions:
(a + a+1 + a+2 + … + b)/(b – a + 1) = (a + b)/2

To solve this equation, we can simplify the left-hand side by multiplying both sides by (b – a + 1):
(a + a+1 + a+2 + … + b) = (b – a + 1)(a + b)/2

Next, we can simplify the right-hand side by distributing:
(a + a+1 + a+2 + … + b) = (a + b)(b – a + 1)/2

Now, we can further simplify the right-hand side by expanding the numerator:
(a + a+1 + a+2 + … + b) = (ab – a^2 + b – ab + a + b)/2

This simplifies to:
(a + a+1 + a+2 + … + b) = (2b – a^2 + 2a)/2

Finally, we can further simplify the equation by canceling out similar terms:
(a + a+1 + a+2 + … + b) = (b – a + 1)

This result shows that the sum of the consecutive integers is equal to the number of integers in the group. This is true for any group of consecutive integers, and it validates the given statement.

In conclusion, the average of a group of consecutive integers is indeed equal to the average of the smallest and largest numbers. This is because the sum of the consecutive integers is equal to the number of integers in the group.

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