To find the four arithmetic means between 15 and 43, we treat the sequence as an arithmetic progression (AP), where 15 is the first term ($a_1 = 15$) and 43 is the last term ($a_6 = 43$). The arithmetic progression consists of 6 terms in total, including the 4 unknown means.

Step 1: Formula for the nth term of an AP

The formula for the nth term of an arithmetic progression is:

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

where:

• $a_n$ is the nth term,
• $a_1$ is the first term,
• $d$ is the common difference, and
• $n$ is the number of terms.

Step 2: Determine the common difference

Here, we know $a_6 = 43$, $a_1 = 15$, and there are 6 terms. Using the formula for $a_6$:

$a_6 = a_1 + (6-1)d$

Substituting the values:

$43 = 15 + 5d$

Solve for $d$:

$43 - 15 = 5d$ $28 = 5d$ $d = \frac{28}{5} = 5.6$

Step 3: Find the arithmetic means

Now that we know $d = 5.6$, we can find the four arithmetic means, which are the second, third, fourth, and fifth terms in the sequence.

• $a_2 = a_1 + d = 15 + 5.6 = 20.6$
• $a_3 = a_1 + 2d = 15 + 2(5.6) = 26.2$
• $a_4 = a_1 + 3d = 15 + 3(5.6) = 31.8$
• $a_5 = a_1 + 4d = 15 + 4(5.6) = 37.4$

Final Answer:

The four arithmetic means between 15 and 43 are: $20.6, 26.2, 31.8, 37.4$

Would you like further details on any part of the solution?

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Here are 5 related questions to extend your understanding:

1. What is the formula for the sum of the first n terms of an arithmetic progression?
2. How would you find the common difference if only the first and last terms of an AP are known?
3. Can an arithmetic progression have a negative common difference? Provide an example.
4. How does the concept of arithmetic means differ from geometric means?
5. How can you find the number of terms in an arithmetic progression given the first and last terms and the common difference?

Tip: In arithmetic progressions, the common difference remains constant between all consecutive terms. This helps in predicting the next terms easily!