If sixth term of a H. P. is `1/61` and its tenth term is `1/105` then the first term of that H.P. is

Options

A. (A) `1/28`
B. (B) `1/39`
C. (C) `1/6`
D. (D) `1/17`

Answer

To solve the problem, we need to find the first term of a Harmonic Progression (H.P.) given that the sixth term is \( \frac{1}{61} \) and the tenth term is \( \frac{1}{105} \).

### Step-by-Step Solution:

1. **Understanding Harmonic Progression**:
- A Harmonic Progression is the reciprocal of an Arithmetic Progression (A.P.). Therefore, if the \( n \)-th term of H.P. is \( H_n \), then the corresponding term in A.P. is \( A_n = \frac{1}{H_n} \).

2. **Setting Up the Terms**:
- Given:
- \( H_6 = \frac{1}{61} \) implies \( A_6 = 61 \)
- \( H_{10} = \frac{1}{105} \) implies \( A_{10} = 105 \)

3. **Using the Formula for A.P.**:
- The formula for the \( n \)-th term of an A.P. is:
\[ A_n = A_1 + (n-1)d \]
- For \( n = 6 \):
\[ A_6 = A_1 + 5d \quad \text{(Equation 1)} \]
- For \( n = 10 \):
\[ A_{10} = A_1 + 9d \quad \text{(Equation 2)} \]

4. **Substituting Known Values**:
- From Equation 1:
\[ 61 = A_1 + 5d \]
- From Equation 2:
\[ 105 = A_1 + 9d \]

5. **Subtracting the Equations**:
- Subtract Equation 1 from Equation 2:
\[ 105 - 61 = (A_1 + 9d) - (A_1 + 5d) \]
\[ 44 = 4d \]
- Solving for \( d \):
\[ d = \frac{44}{4} = 11 \]

6. **Finding \( A_1 \)**:
- Substitute \( d \) back into Equation 1:
\[ 61 = A_1 + 5 \times 11 \]
\[ 61 = A_1 + 55 \]
\[ A_1 = 61 - 55 = 6 \]

7. **Finding the First Term of H.P.**:
- Since the first term of H.P. is the reciprocal of the first term of A.P.:
\[ H_1 = \frac{1}{A_1} = \frac{1}{6} \]

### Final Answer:
The first term of the Harmonic Progression is \( \frac{1}{6} \).