A. Give what is asked.

1. The 10 term of the arithmetic sequence if a₁ = -15 and d = 6.

2. The 39 term of the arithmetic sequence if a₁ = 40 and d = 12.

B. Find the specified term of each arithmetic sequence.

1. 1.4, 4.5, 7.6, ...; the 41. term

2. 9, 18, 27,...; the 23 term

C. Insert the specified number of arithmetic means between the two given arithmetic extremes.

1. Three arithmetic means between 2 and 22.

2. Four arithmetic means between 8 and 23.

Let's solve each of these problems step by step.

### Part A: Finding the specific terms of an arithmetic sequence

1. **Find the 10th term of the arithmetic sequence where \( a_1 = -15 \) and \( d = 6 \):**

   The general formula for the \( n \)th term of an arithmetic sequence is:
   \[
   a_n = a_1 + (n - 1) \times d
   \]
   For the 10th term (\( n = 10 \)):
   \[
   a_{10} = -15 + (10 - 1) \times 6 = -15 + 9 \times 6 = -15 + 54 = 39
   \]
   So, the 10th term is **39**.

2. **Find the 39th term of the arithmetic sequence where \( a_1 = 40 \) and \( d = 12 \):**

   Using the same formula:
   \[
   a_{39} = 40 + (39 - 1) \times 12 = 40 + 38 \times 12 = 40 + 456 = 496
   \]
   So, the 39th term is **496**.

### Part B: Finding the specified term of each arithmetic sequence

1. **Find the 41st term of the sequence \( 1.4, 4.5, 7.6, \ldots \):**

   First, determine the common difference \( d \):
   \[
   d = 4.5 - 1.4 = 3.1
   \]
   Now, use the formula for the 41st term (\( n = 41 \)):
   \[
   a_{41} = 1.4 + (41 - 1) \times 3.1 = 1.4 + 40 \times 3.1 = 1.4 + 124 = 125.4
   \]
   So, the 41st term is **125.4**.

2. **Find the 23rd term of the sequence \( 9, 18, 27, \ldots \):**

   Here, the common difference \( d \) is:
   \[
   d = 18 - 9 = 9
   \]
   Now, use the formula for the 23rd term (\( n = 23 \)):
   \[
   a_{23} = 9 + (23 - 1) \times 9 = 9 + 22 \times 9 = 9 + 198 = 207
   \]
   So, the 23rd term is **207**.

### Part C: Inserting arithmetic means between given extremes

1. **Insert three arithmetic means between 2 and 22:**

   If there are 3 arithmetic means, there are 5 terms in total: \( a_1 = 2 \), \( a_5 = 22 \).
   The formula for the \( n \)th term gives:
   \[
   a_5 = a_1 + 4d = 2 + 4d = 22
   \]
   Solve for \( d \):
   \[
   4d = 20 \implies d = 5
   \]
   The sequence is: \( 2, 7, 12, 17, 22 \).

   So, the three arithmetic means are **7, 12,** and **17**.

2. **Insert four arithmetic means between 8 and 23:**

   If there are 4 arithmetic means, there are 6 terms in total: \( a_1 = 8 \), \( a_6 = 23 \).
   The formula for the \( n \)th term gives:
   \[
   a_6 = a_1 + 5d = 8 + 5d = 23
   \]
   Solve for \( d \):
   \[
   5d = 15 \implies d = 3
   \]
   The sequence is: \( 8, 11, 14, 17, 20, 23 \).

   So, the four arithmetic means are **11, 14, 17,** and **20**.

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Would you like further details or have any questions? 

Here are some related questions:
1. How do you derive the formula for the nth term of an arithmetic sequence?
2. What happens if the common difference \( d \) is negative?
3. How can you determine the sum of the first \( n \) terms in an arithmetic sequence?
4. How does the formula change if you know the last term of the sequence?
5. Can an arithmetic sequence have a non-integer common difference?
6. What is the geometric interpretation of an arithmetic sequence?
7. How can you find the middle term in an arithmetic sequence with an odd number of terms?
8. What are some real-world applications of arithmetic sequences?

**Tip:** Always check the sequence’s common difference to ensure it’s consistent before applying formulas.

Explore solutions to arithmetic sequence problems including finding specific terms like the 10th and 39th terms, and inserting arithmetic means between given extremes. Includes step-by-step calculations and formulas used. Suitable for high school students.

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