Solving the Consecutive Integer Average Problem

The question asks us to find the largest integer in a set of 18 consecutive integers whose average is 22.5. Understanding the properties of consecutive integers and averages is key here.

Understanding Average of Consecutive Integers

For any set of consecutive integers:

• If the count of integers is odd, the average is the middle integer.
• If the count of integers is even, the average is exactly halfway between the two middle integers.

In this problem, we have 18 consecutive integers, which is an even count. The given average is 22.5.

Finding the Middle Integers

Since the count is 18 (even), the average 22.5 is exactly halfway between the 9th and 10th integers in the sorted set. The two integers that are consecutive and have 22.5 as their average are 22 and 23.

Check: $\frac{22 + 23}{2} = \frac{45}{2} = 22.5$.

So, the 9th integer in the set is 22, and the 10th integer is 23.

Finding the Largest Integer

We know the 10th integer is 23. The set has 18 integers in total. The largest integer is the 18th integer in the set.

To get from the 10th integer to the 18th integer, we need to move forward $18 - 10 = 8$ positions. Since the integers are consecutive, each step forward adds 1 to the value.

Largest integer = (Value of 10th integer) + (Number of steps forward)

Largest integer = $23 + 8 = 31$.

Finding the Smallest Integer (Optional but helpful)

We know the 9th integer is 22. To get from the 9th integer to the 1st integer, we need to move backward $9 - 1 = 8$ positions. Since the integers are consecutive, each step backward subtracts 1 from the value.

Smallest integer = (Value of 9th integer) - (Number of steps backward)

Smallest integer = $22 - 8 = 14$.

So, the set of 18 consecutive integers starts at 14 and ends at 31: $\{14, 15, \dots, 30, 31\}$.

Verification

The set is $\{14, 15, \dots, 31\}$. There are $31 - 14 + 1 = 18$ integers.

The smallest is 14, and the largest is 31.

The average of an arithmetic series (like consecutive integers) is also the average of the first and last term:

Average = $\frac{\text{First Term} + \text{Last Term}}{2} = \frac{14 + 31}{2} = \frac{45}{2} = 22.5$.

This matches the given average, confirming our calculations.

The largest integer in the set is 31.

Revision Table: Consecutive Integers and Average

Additional Information: Finding Consecutive Integers from Average

If you know the average and the count (N) of consecutive integers:

1. If N is odd, the middle term is the average. You can find the terms before and after by subtracting/adding 1 repeatedly.
2. If N is even, the average is halfway between the two middle terms. The two middle terms are (Average - 0.5) and (Average + 0.5). In our case, 22.5 - 0.5 = 22 and 22.5 + 0.5 = 23. These are the two middle integers.

Once you have the middle term(s), you can determine their position (e.g., for 18 integers, the middle are 9th and 10th). Then, calculate the first and last terms by moving backward and forward from the middle term(s) based on their position.

For N integers, the middle position(s) are around $(N+1)/2$. If N is even, the positions are $N/2$ and $N/2 + 1$. For N=18, positions are $18/2 = 9$ and $18/2 + 1 = 10$.