The problem asks for the sum of the smallest and largest numbers in a group of 8 consecutive even numbers, given that their average is 41.

Key Property of Consecutive Numbers

For any arithmetic sequence (like consecutive even numbers), the average is equal to the average of the first and last term. That is:

$ \text{Average} = \frac{\text{Smallest Number} + \text{Largest Number}}{2} $

Calculating the Sum

We are given that the average of the 8 consecutive even numbers is 41.

Using the property mentioned above:

$ 41 = \frac{\text{Smallest Number} + \text{Largest Number}}{2} $

To find the sum of the smallest and largest numbers, we multiply the average by 2:

$ \text{Smallest Number} + \text{Largest Number} = 41 \times 2 $

$ \text{Smallest Number} + \text{Largest Number} = 82 $

Verification (Optional)

Since there are 8 consecutive even numbers, the average (41) lies between the 4th and 5th numbers. The two middle even numbers must be 40 and 42.

• 4th Number = 40
• 5th Number = 42

Working backwards and forwards:

• The 8 numbers are: 34, 36, 38, 40, 42, 44, 46, 48.
• Smallest Number = 34
• Largest Number = 48
• Sum = $34 + 48 = 82$.

The sum of the smallest and largest numbers is 82.