Question

The arithmetic mean of 10 consecutive even integers is 3. What is the least of these 10 even integers?

Answer 1

Answer:

-6

Step-by-step explanation:

2n can be the smallest integer, and 2n + 18 will be the largest integer.

The sum of this, divided by two, will result in the average/mean.

(2n + 2n + 18)/2 = 3

Multiply each side by 2:

(2n + 2n + 18)/2 ⋅ 2 = 3 ⋅ 2

2n + 2n + 18  =  6

Combine the like terms:

4n + 18 = 6

Subtract 18 from both sides:

4n + 18 - 18 = 6 - 18

4n = -12

Divide each side by 4:

4n/4 = -12/4

n = -3

Since we decided to go by 2n:

2n = 2(-3) = -6