If the average (arithmetic mean) of 5 consecutive integers is 12, what is the sum of the least and greatest of the 5 integers?
The correct answer and explanation is:
Correct Answer: 24
To solve the problem, begin by understanding the nature of consecutive integers and their average.
Let the five consecutive integers be represented as: x−2,x−1,x,x+1,x+2x – 2, x – 1, x, x + 1, x + 2
These are five numbers in a row, centered around the middle number xx.
The average (arithmetic mean) of these five integers is given as 12. Since the numbers are symmetric around the middle value, the average of these five integers is exactly the middle number, which is xx. Therefore: x=12x = 12
So the five integers are: 12−2=10,12−1=11,12,12+1=13,12+2=1412 – 2 = 10,\quad 12 – 1 = 11,\quad 12,\quad 12 + 1 = 13,\quad 12 + 2 = 14
The least integer is 10 and the greatest integer is 14.
Their sum is: 10+14=2410 + 14 = 24
This is the final answer.
Explanation:
The key idea is that in any list of an odd number of consecutive integers, the average is always the middle number. That is why representing the five integers as x−2x – 2 to x+2x + 2 simplifies the process. Since the question provides the average as 12, it means the middle number of this sequence is 12. Once the integers are determined, it is straightforward to find the least and greatest, then add them together.
This method is efficient and avoids unnecessary steps like computing the entire sum and dividing it again. Using symmetry in arithmetic sequences provides a quick path to the solution. The final result is 24, which is the sum of the smallest and largest of the five consecutive integers.