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The problem asks to compute the 16th term of a Harmonic Progression (HP), given that the 6th and 11th terms of the HP are 10 and 18, respectively.

### Step-by-Step Explanation:

1. **Understanding the Relationship Between HP and AP**:
   In a Harmonic Progression (HP), the reciprocals of the terms form an Arithmetic Progression (AP). If the terms of an HP are \( a_1, a_2, a_3, \ldots \), then the reciprocals \( \frac{1}{a_1}, \frac{1}{a_2}, \frac{1}{a_3}, \ldots \) form an AP.

   Let's assume the terms of the HP are \( a_n \), and the corresponding AP is represented as \( b_n \), where \( b_n = \frac{1}{a_n} \).

2. **Given Information**:
   - The 6th term of the HP: \( a_6 = 10 \) ⟹ \( b_6 = \frac{1}{10} \)
   - The 11th term of the HP: \( a_{11} = 18 \) ⟹ \( b_{11} = \frac{1}{18} \)

3. **General Formula for the Arithmetic Progression (AP)**:
   The general formula for the \(n\)-th term of an AP is:
   \[
   b_n = b_1 + (n - 1) d
   \]
   where \( b_1 \) is the first term of the AP, and \( d \) is the common difference.

4. **Setting Up Two Equations for \( b_6 \) and \( b_{11} \)**:
   Using the formula for the AP:

   - For the 6th term:
     \[
     b_6 = b_1 + (6 - 1) d = b_1 + 5d = \frac{1}{10}
     \]
   - For the 11th term:
     \[
     b_{11} = b_1 + (11 - 1) d = b_1 + 10d = \frac{1}{18}
     \]

5. **Solving the System of Equations**:
   We now have two equations:
   \[
   b_1 + 5d = \frac{1}{10} \quad \text{(1)}
   \]
   \[
   b_1 + 10d = \frac{1}{18} \quad \text{(2)}
   \]
   Subtract equation (1) from equation (2):
   \[
   (b_1 + 10d) - (b_1 + 5d) = \frac{1}{18} - \frac{1}{10}
   \]
   Simplifying:
   \[
   5d = \frac{1}{18} - \frac{1}{10}
   \]
   Find a common denominator for the fractions:
   \[
   5d = \frac{5}{90} - \frac{9}{90} = \frac{-4}{90} = \frac{-2}{45}
   \]
   Therefore:
   \[
   d = \frac{-2}{45 \times 5} = \frac{-2}{225}
   \]

6. **Substitute \( d \) Back into One of the Equations**:
   Substitute \( d = \frac{-2}{225} \) into equation (1):
   \[
   b_1 + 5 \times \frac{-2}{225} = \frac{1}{10}
   \]
   \[
   b_1 - \frac{10}{225} = \frac{1}{10}
   \]
   \[
   b_1 = \frac{1}{10} + \frac{2}{45} = \frac{9}{90} + \frac{4}{90} = \frac{13}{90}
   \]

7. **Finding the 16th Term**:
   Now that we have \( b_1 = \frac{13}{90} \) and \( d = \frac{-2}{225} \), we can find the 16th term of the AP, which corresponds to the reciprocal of the 16th term of the HP:
   \[
   b_{16} = b_1 + (16 - 1) d = \frac{13}{90} + 15 \times \frac{-2}{225}
   \]
   \[
   b_{16} = \frac{13}{90} + \frac{-30}{225}
   \]
   Convert \( \frac{-30}{225} \) to have a denominator of 90:
   \[
   b_{16} = \frac{13}{90} + \frac{-12}{90} = \frac{1}{90}
   \]

8. **Conclusion**:
   The 16th term of the HP, \( a_{16} \), is the reciprocal of \( b_{16} \):
   \[
   a_{16} = \frac{1}{b_{16}} = \frac{1}{\frac{1}{90}} = 90
   \]

Thus, the 16th term of the harmonic progression is **90**.

Would you like further details on this process? Here are some related questions:

1. How is an arithmetic progression different from a harmonic progression?
2. Can you explain how the reciprocal relationship works between HP and AP?
3. How do you determine the common difference in an arithmetic progression?
4. What would happen if the given terms were part of a geometric progression instead?
5. How does solving systems of equations help in progressions?

**Tip:** Always remember that for a Harmonic Progression, the reciprocals of the terms follow an Arithmetic Progression!

Compute the 16th term of a Harmonic Progression if the 6th and 11th terms are 10 and 18, respectively.

This problem involves finding the 16th term of a Harmonic Progression when the 6th and 11th terms are given as 10 and 18. Learn the step-by-step solution using the reciprocal relationship between harmonic and arithmetic progressions.

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