To insert three arithmetic means between 16 and -8, let's analyze the problem.

Given:

• First term (a₁) = 16
• Last term (a₅) = -8
• Number of terms (n) = 5

In an arithmetic sequence, the common difference $d$ is constant between consecutive terms. The nth term of an arithmetic sequence is given by the formula:

$a_n = a_1 + (n-1) \cdot d$

For this sequence: $a_5 = a_1 + 4d$

Substituting the known values: $-8 = 16 + 4d$

Solving for $d$: $4d = -8 - 16$ $4d = -24$ $d = -6$

So the common difference $d$ is -6.

Now, let's find the missing terms:

1. Second term: $a_2 = 16 + (-6) = 10$
2. Third term: $a_3 = 10 + (-6) = 4$
3. Fourth term: $a_4 = 4 + (-6) = -2$

The complete sequence is:

16, 10, 4, -2, -8

Would you like more details or have any questions? Here are 5 related questions:

1. How is the formula for the nth term of an arithmetic sequence derived?
2. What if you had to find 4 arithmetic means instead of 3?
3. How do you determine if a sequence is arithmetic?
4. Can arithmetic sequences be represented graphically?
5. What happens if the common difference is positive instead of negative?

Tip: In arithmetic sequences, the common difference can be found by subtracting any term from the next term.