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.

Learn how to insert arithmetic means in a sequence with a detailed step-by-step solution. This problem covers arithmetic sequences, including finding the common difference and determining missing terms. Suitable for students in Grades 7-9.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation