The average of 29 consecutive even integers is 60. What is the largest (highest) integer in this sequence?

Introduction / Context:
This problem tests your understanding of averages and properties of consecutive even integers. When you have an odd number of equally spaced values in a sequence (such as consecutive even integers), the average is equal to the middle term of the sequence. Here, the average of 29 consecutive even integers is given as 60, and you are asked to find the largest integer in the sequence. Recognizing the symmetry of such sequences makes this question much easier.

Given Data / Assumptions:

- There are 29 consecutive even integers. - Their average is 60. - The integers are evenly spaced with a common difference of 2. - We must determine the largest integer in the sequence.

Concept / Approach:
For any arithmetic progression with an odd number of terms, the average of the terms equals the middle term. Consecutive even integers form such an arithmetic progression with common difference 2. With 29 terms, the middle term is the 15th term. Since the average is 60, the 15th term is 60. Once we know the middle term, we can move forward or backward using the common difference to find any other term, including the largest, which is the 29th term in this case.

Step-by-Step Solution:
Step 1: Recognize that 29 is odd, so there is a single middle term in the sequence. Step 2: The number of terms is 29, so the middle term is the (29 + 1) / 2 = 15th term. Step 3: For an arithmetic progression, the average of all terms equals the middle term when the number of terms is odd. Step 4: Therefore, the 15th term is equal to the average, which is 60. Step 5: The sequence consists of consecutive even integers with common difference 2. So each step to the right adds 2. Step 6: The largest integer in the sequence is the 29th term. Step 7: The 29th term is 14 places after the 15th term, because 29 - 15 = 14. Step 8: Each step adds 2, so the increment from the 15th to the 29th term is 14 * 2 = 28. Step 9: Add this increment to the middle term: 60 + 28 = 88. Step 10: Hence, the largest integer in the sequence is 88.

Verification / Alternative check:
You can also construct part of the sequence around 60. The terms around the middle are 60 - 28, 60 - 26, ..., 60, ..., 60 + 26, 60 + 28. Counting 14 even steps below 60 and 14 even steps above gives 29 terms in total, with the highest being 60 + 28 = 88. This direct reasoning confirms the earlier calculation and shows that the structure is symmetric around 60.

Why Other Options Are Wrong:
Values 118, 176 and 120 correspond to using incorrect differences or misinterpreting how many steps away from the middle term the last term is. For example, taking 29 * 2 added to 60 instead of 14 * 2 leads to a much larger, incorrect value. Only 88 preserves the correct distance from the middle term in a set of 29 evenly spaced numbers.

Common Pitfalls:
Some learners forget that the average equals the middle term only when the number of terms is odd. Others miscount the number of steps between the middle and the last term, or mistakenly use a common difference of 1 instead of 2 for even integers. Always double check the step count and the common difference when moving from the middle to the endpoints in such sequences.

Final Answer:
The highest integer in the sequence is 88, which matches option A.