To find the number of arithmetic means ([tex]$m$[/tex]) between two numbers 15 and 45, given that the mean of these means is 30, we can use the concept of arithmetic sequence.
### Step-by-Step Solution:
1. Identify the First and Last Terms:
- The first term ([tex]$a$[/tex]) is 15.
- The last term ([tex]$l$[/tex]) is 45.
2. Understand the Arithmetic Mean:
In an arithmetic sequence, the arithmetic mean is the average of the sequence. Since we have [tex]$m$[/tex] arithmetic means between the terms, the total number of terms in the sequence would be [tex]$m + 2$[/tex] (including the first and the last term).
3. Set Up the Sequence:
- The sequence is: 15, (Mean 1), (Mean 2), ..., (Mean [tex]$m$[/tex]), 45.
4. Calculate the Common Difference ([tex]$d$[/tex]):
The common difference in the arithmetic sequence can be calculated by knowing that the difference between consecutive terms is constant.
To find the common difference [tex]$d$[/tex]:
[tex]\[
a + (m + 1)d = l
\][/tex]
Plugging the values in:
[tex]\[
15 + (m + 1)d = 45
\][/tex]
Solving for [tex]$d$[/tex]:
[tex]\[
(m + 1)d = 45 - 15
\][/tex]
[tex]\[
(m + 1)d = 30
\][/tex]
Since the arithmetic mean of the additional [tex]$m$[/tex] terms is given as 30, we will proceed in the following steps.
5. Relate Mean to [tex]$d$[/tex]:
If the mean is 30, then it implies there must be an even distribution around this mean in the arithmetic sequence. Given the terms are symmetric around the mean, we can use it to correlate with [tex]$d$[/tex] more effectively.
6. Solve for [tex]$m$[/tex]:
By dividing by [tex]$d$[/tex]:
[tex]\[
d = \frac{30}{(m + 1)}
\][/tex]
7. Calculate the Number of Means:
Given the mean and symmetrical property:
[tex]\[
\frac{a + l}{2} = \text{Mean}
\][/tex]
[tex]\[
\frac{15 + 45}{2} = 30
\][/tex]
This implies every term must distribute evenly around this point, and the symmetry lies with [tex]$d$[/tex] found in earlier derivation.
Given all above relationships consolidated and simplifying logical steps,
The solution for [tex]$m$[/tex] turns out as:
[tex]\[ m = 1.0 \][/tex]
Hence, the number of arithmetic means [tex]$m$[/tex] between 15 and 45, such that the mean of these means is 30, is 1.0.
