Math Problem Statement

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.

Solution

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 a1=15a_1 = -15 and d=6d = 6:

    The general formula for the nnth term of an arithmetic sequence is: an=a1+(n1)×da_n = a_1 + (n - 1) \times d For the 10th term (n=10n = 10): a10=15+(101)×6=15+9×6=15+54=39a_{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 a1=40a_1 = 40 and d=12d = 12:

    Using the same formula: a39=40+(391)×12=40+38×12=40+456=496a_{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 1.4,4.5,7.6,1.4, 4.5, 7.6, \ldots:

    First, determine the common difference dd: d=4.51.4=3.1d = 4.5 - 1.4 = 3.1 Now, use the formula for the 41st term (n=41n = 41): a41=1.4+(411)×3.1=1.4+40×3.1=1.4+124=125.4a_{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 9,18,27,9, 18, 27, \ldots:

    Here, the common difference dd is: d=189=9d = 18 - 9 = 9 Now, use the formula for the 23rd term (n=23n = 23): a23=9+(231)×9=9+22×9=9+198=207a_{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: a1=2a_1 = 2, a5=22a_5 = 22. The formula for the nnth term gives: a5=a1+4d=2+4d=22a_5 = a_1 + 4d = 2 + 4d = 22 Solve for dd: 4d=20    d=54d = 20 \implies d = 5 The sequence is: 2,7,12,17,222, 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: a1=8a_1 = 8, a6=23a_6 = 23. The formula for the nnth term gives: a6=a1+5d=8+5d=23a_6 = a_1 + 5d = 8 + 5d = 23 Solve for dd: 5d=15    d=35d = 15 \implies d = 3 The sequence is: 8,11,14,17,20,238, 11, 14, 17, 20, 23.

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


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 dd is negative?
  3. How can you determine the sum of the first nn 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.

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Math Problem Analysis

Mathematical Concepts

Arithmetic Sequences
Nth Term Formula
Inserting Arithmetic Means

Formulas

Nth term of an arithmetic sequence

Theorems

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Suitable Grade Level

High School