To insert 4 Harmonic Means (H.M.) between \( \frac{3}{4} \) and \( \frac{3}{19} \), we can follow these steps:
### Step-by-Step Solution:
1. **Understanding Harmonic Means**:
The Harmonic Mean (H.M.) of two numbers \( a \) and \( b \) is given by the formula:
\[
H.M. = \frac{2ab}{a + b}
\]
However, in this case, we will use the property that the harmonic means can be found by taking the reciprocals of the numbers and forming an Arithmetic Progression (A.P.).
2. **Setting up the Inverses**:
Let the H.M.s be \( H_1, H_2, H_3, H_4 \). The corresponding inverses will be:
\[
\frac{4}{3}, \frac{1}{H_1}, \frac{1}{H_2}, \frac{1}{H_3}, \frac{1}{H_4}, \frac{19}{3}
\]
These values will be in A.P.
3. **Identifying the First and Last Terms**:
The first term \( a = \frac{4}{3} \) and the last term \( l = \frac{19}{3} \).
4. **Finding the Common Difference**:
Since there are 6 terms in total (including the first and last), we can express the last term in terms of the first term and the common difference \( d \):
\[
\frac{19}{3} = \frac{4}{3} + 5d
\]
Rearranging gives:
\[
5d = \frac{19}{3} - \frac{4}{3} = \frac{15}{3} = 5
\]
Thus, we find:
\[
d = 1
\]
5. **Finding Each H.M.**:
- For \( H_1 \):
\[
\frac{1}{H_1} = \frac{4}{3} + d = \frac{4}{3} + 1 = \frac{4}{3} + \frac{3}{3} = \frac{7}{3}
\]
Therefore,
\[
H_1 = \frac{3}{7}
\]
- For \( H_2 \):
\[
\frac{1}{H_2} = \frac{4}{3} + 2d = \frac{4}{3} + 2 = \frac{4}{3} + \frac{6}{3} = \frac{10}{3}
\]
Therefore,
\[
H_2 = \frac{3}{10}
\]
- For \( H_3 \):
\[
\frac{1}{H_3} = \frac{4}{3} + 3d = \frac{4}{3} + 3 = \frac{4}{3} + \frac{9}{3} = \frac{13}{3}
\]
Therefore,
\[
H_3 = \frac{3}{13}
\]
- For \( H_4 \):
\[
\frac{1}{H_4} = \frac{4}{3} + 4d = \frac{4}{3} + 4 = \frac{4}{3} + \frac{12}{3} = \frac{16}{3}
\]
Therefore,
\[
H_4 = \frac{3}{16}
\]
6. **Final Result**:
The four Harmonic Means between \( \frac{3}{4} \) and \( \frac{3}{19} \) are:
\[
H_1 = \frac{3}{7}, \quad H_2 = \frac{3}{10}, \quad H_3 = \frac{3}{13}, \quad H_4 = \frac{3}{16}
\]
