The image contains a set of problems where you need to find arithmetic means between two numbers. Here is a transcription of the problems:

1. Three arithmetic means between 5 and 25.
2. Four arithmetic means between 8 and -12.
3. Three arithmetic means between 4 and 16.
4. Five arithmetic means between 30 and 6.
5. Four arithmetic means between 6 and 41.

Approach to solve these:

The arithmetic means are the terms in an arithmetic sequence between two given numbers, where the difference between consecutive terms is constant. The steps are:

1. Let the sequence include the given numbers and the arithmetic means.
2. Use the formula for the $n$-th term of an arithmetic sequence:
   $a_n = a_1 + (n-1) \cdot d$ Where:
   • $a_n$ is the $n$-th term
   • $a_1$ is the first term
   • $d$ is the common difference

Let’s solve each of these problems using this method.

1. Three arithmetic means between 5 and 25

• Total terms in the sequence: 5 (including 5 and 25).
• First term $a_1 = 5$, fifth term $a_5 = 25$.
• Use the formula for the 5th term: $25 = 5 + (5-1) \cdot d \implies 25 = 5 + 4d$ Solving for $d$: $4d = 20 \implies d = 5$ Now, the sequence is: 5, 10, 15, 20, 25. The arithmetic means are 10, 15, 20.

2. Four arithmetic means between 8 and -12

• Total terms: 6 (including 8 and -12).
• First term $a_1 = 8$, sixth term $a_6 = -12$.
• Use the formula for the 6th term: $-12 = 8 + (6-1) \cdot d \implies -12 = 8 + 5d$ Solving for $d$: $5d = -20 \implies d = -4$ Now, the sequence is: 8, 4, 0, -4, -8, -12. The arithmetic means are 4, 0, -4, -8.

3. Three arithmetic means between 4 and 16

• Total terms: 5 (including 4 and 16).
• First term $a_1 = 4$, fifth term $a_5 = 16$.
• Use the formula for the 5th term: $16 = 4 + (5-1) \cdot d \implies 16 = 4 + 4d$ Solving for $d$: $4d = 12 \implies d = 3$ Now, the sequence is: 4, 7, 10, 13, 16. The arithmetic means are 7, 10, 13.

4. Five arithmetic means between 30 and 6

• Total terms: 7 (including 30 and 6).
• First term $a_1 = 30$, seventh term $a_7 = 6$.
• Use the formula for the 7th term: $6 = 30 + (7-1) \cdot d \implies 6 = 30 + 6d$ Solving for $d$: $6d = -24 \implies d = -4$ Now, the sequence is: 30, 26, 22, 18, 14, 10, 6. The arithmetic means are 26, 22, 18, 14, 10.

5. Four arithmetic means between 6 and 41

• Total terms: 6 (including 6 and 41).
• First term $a_1 = 6$, sixth term $a_6 = 41$.
• Use the formula for the 6th term: $41 = 6 + (6-1) \cdot d \implies 41 = 6 + 5d$ Solving for $d$: $5d = 35 \implies d = 7$ Now, the sequence is: 6, 13, 20, 27, 34, 41. The arithmetic means are 13, 20, 27, 34.

Would you like more details on any of these solutions or help with related topics?

Related Questions:

1. How do you find the common difference in an arithmetic sequence if only two terms are given?
2. How can you find the number of terms in an arithmetic sequence given the first and last term?
3. What is the general formula for the sum of an arithmetic series?
4. How would the method change if you were asked to find geometric means instead of arithmetic means?
5. Can you apply arithmetic sequences in real-life situations, like finance or physics?

Tip:

For arithmetic sequences, the common difference is key. Once you have that, generating terms becomes straightforward. Practice identifying patterns in sequences!