The average of 19 consecutive even integers is 50. What is the value of the highest integer in this sequence?

Introduction / Context:
This question tests understanding of averages and properties of consecutive even integers. When numbers are consecutive and equally spaced, the average is equal to the middle term of the sequence. Once we identify the middle term, we can easily move up or down by the common difference to reach the highest or lowest term in the list.

Given Data / Assumptions:
- We have 19 consecutive even integers.
- The average of these 19 integers is 50.
- The integers are equally spaced with a common difference of 2 because they are even and consecutive.
- We are asked to find the highest integer in this sequence.

Concept / Approach:
For any arithmetic progression, including consecutive even integers, the average of all terms is equal to the middle term when the number of terms is odd. Since we have 19 terms, the 10th term in the ordered list will be the middle term and equal to the average. Then the highest term will be 9 steps above the middle term, each step increasing by 2, because there are 9 integers after the middle term in a list of 19 terms.

Step-by-Step Solution:
Step 1: Note that the number of terms is 19, which is odd, so the middle term is the 10th term.Step 2: For an arithmetic progression with an odd number of terms, the average equals the middle term.Step 3: Therefore the 10th term is equal to the average, which is 50.Step 4: There are 9 numbers after the 10th term, and each successive term increases by 2.Step 5: Highest term = 10th term + 9 × 2 = 50 + 18 = 68.

Verification / Alternative check:
You can list the sequence around the middle term to verify: 10th term is 50, so the sequence looks like 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68. The highest term is clearly 68. If you compute the average of these 19 terms, you again obtain 50, which confirms the method.

Why Other Options Are Wrong:
Values such as 88, 100, or 126 are far larger than the correct upper term of a sequence centered at 50 with only 19 even numbers and would require many more than 9 steps of size 2 from the middle term. The value 72 is closer but would correspond to 11 steps above 50, implying more than 19 terms, so it does not match the given length of the sequence.

Common Pitfalls:
Some learners try to build the sequence from an assumed lowest term without using the property that the average equals the middle term. Others mistakenly add 19 × 2 to the average, which is incorrect because that would count too many steps. Remember that for an odd number of equally spaced values, the average is always the central term, which greatly simplifies problems like this.

Final Answer:
The highest integer in the sequence is 68.