In a H.P. if `5^(th)` term is 6 and `3^(rd)` term is 10. Find the `2^(nd)` term

Answer

To solve the problem, we need to find the second term of a Harmonic Progression (H.P.) given that the 5th term is 6 and the 3rd term is 10.

### Step-by-Step Solution:

1. **Understanding Harmonic Progression (H.P.)**:
- In H.P., if the terms 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 Arithmetic Progression (A.P.).

2. **Setting Up the Terms**:
- Let the first term of the H.P. be \( a \) and the common difference of the corresponding A.P. be \( d \).
- The 5th term of the H.P. can be expressed as:
\[ T_5 = \frac{1}{a + 4d} \]
- The 3rd term of the H.P. can be expressed as:
\[ T_3 = \frac{1}{a + 2d} \]

3. **Using Given Information**:
- We know that:
\[ T_5 = 6 \implies \frac{1}{a + 4d} = 6 \implies a + 4d = \frac{1}{6} \quad \text{(Equation 1)} \]
\[ T_3 = 10 \implies \frac{1}{a + 2d} = 10 \implies a + 2d = \frac{1}{10} \quad \text{(Equation 2)} \]

4. **Solving the Equations**:
- From Equation 1:
\[ a + 4d = \frac{1}{6} \]
- From Equation 2:
\[ a + 2d = \frac{1}{10} \]
- Now, subtract Equation 2 from Equation 1:
\[ (a + 4d) - (a + 2d) = \frac{1}{6} - \frac{1}{10} \]
\[ 2d = \frac{1}{6} - \frac{1}{10} \]

5. **Finding a Common Denominator**:
- The least common multiple of 6 and 10 is 30:
\[ \frac{1}{6} = \frac{5}{30}, \quad \frac{1}{10} = \frac{3}{30} \]
- Therefore:
\[ 2d = \frac{5}{30} - \frac{3}{30} = \frac{2}{30} = \frac{1}{15} \]
- Thus:
\[ d = \frac{1}{30} \]

6. **Substituting Back to Find \( a \)**:
- Substitute \( d \) back into Equation 2:
\[ a + 2\left(\frac{1}{30}\right) = \frac{1}{10} \]
\[ a + \frac{2}{30} = \frac{3}{30} \]
\[ a = \frac{3}{30} - \frac{2}{30} = \frac{1}{30} \]

7. **Finding the 2nd Term**:
- The 2nd term of the H.P. is given by:
\[ T_2 = \frac{1}{a + 1d} \]
- Substitute \( a \) and \( d \):
\[ T_2 = \frac{1}{\frac{1}{30} + \frac{1}{30}} = \frac{1}{\frac{2}{30}} = \frac{30}{2} = 15 \]

### Final Answer:
The 2nd term of the H.P. is **15**.