Insert 3 arithmetic mean between 5 and 30


To find three arithmetic means between 5 and 30, we'll assume that they form an arithmetic progression (AP). In an AP, the difference between consecutive terms is constant.

Given:
- The first term (\(a_1\)) = 5
- The fifth term (\(a_5\)) = 30
- We are looking for the second (\(a_2\)), third (\(a_3\)), and fourth (\(a_4\)) terms.

Let the common difference be \(d\). Using the formula for the \(n\)-th term of an arithmetic progression:

\[
a_n = a_1 + (n-1) \cdot d
\]

For the fifth term \(a_5 = 30\), we substitute the known values:

\[
30 = 5 + (5-1) \cdot d
\]
\[
30 = 5 + 4d
\]
\[
4d = 25
\]
\[
d = \frac{25}{4} = 6.25
\]

Now that we know the common difference \(d = 6.25\), we can find the three arithmetic means:

\[
a_2 = a_1 + d = 5 + 6.25 = 11.25
\]
\[
a_3 = a_1 + 2d = 5 + 2(6.25) = 17.5
\]
\[
a_4 = a_1 + 3d = 5 + 3(6.25) = 23.75
\]

Thus, the three arithmetic means between 5 and 30 are:

\[
11.25, 17.5, 23.75
\]

Would you like more details or have any questions about the process?

Here are 5 related questions:
1. How do you find the common difference in an arithmetic progression?
2. What is the formula for the sum of an arithmetic progression?
3. How does an arithmetic progression differ from a geometric progression?
4. Can you find 4 arithmetic means between two given numbers?
5. How is the general term of an arithmetic progression derived?

**Tip:** In arithmetic progressions, the common difference \(d\) stays the same between any two consecutive terms.

Insert 3 arithmetic mean between 5 and 30

Learn how to insert 3 arithmetic means between 5 and 30 using the arithmetic progression formula. This problem explains how to calculate the common difference and find the intermediate terms in a step-by-step process. Ideal for students in Grades 8-10.

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