In a H.P., `T_(7) = (1)/(10) and T_(12) = (1)/(25) "then" T_(20)` is

Options

A. `(1)/(37)`
B. `(1)/(40)`
C. `(1)/(49)`
D. `(1)/(45)`

Answer

To solve the problem, we need to find the 20th term \( T_{20} \) of a Harmonic Progression (H.P.) given that \( T_7 = \frac{1}{10} \) and \( T_{12} = \frac{1}{25} \).

### Step-by-Step Solution:

1. **Understanding the terms of H.P.**:
In a Harmonic Progression, the \( n \)-th term \( T_n \) can be expressed in terms of the first term \( A \) and the common difference \( D \) of the corresponding Arithmetic Progression (A.P.) as follows:
\[ T_n = \frac{1}{A + (n-1)D} \]

2. **Setting up the equations**:
From the given information:
- For \( T_7 \):
\[ T_7 = \frac{1}{A + (7-1)D} = \frac{1}{A + 6D} = \frac{1}{10} \]
This gives us the equation:
\[ A + 6D = 10 \quad \text{(Equation 1)} \]

- For \( T_{12} \):
\[ T_{12} = \frac{1}{A + (12-1)D} = \frac{1}{A + 11D} = \frac{1}{25} \]
This gives us the equation:
\[ A + 11D = 25 \quad \text{(Equation 2)} \]

3. **Solving the equations**:
Now we have two equations:
- \( A + 6D = 10 \) (Equation 1)
- \( A + 11D = 25 \) (Equation 2)

We can subtract Equation 1 from Equation 2:
\[ (A + 11D) - (A + 6D) = 25 - 10 \]
Simplifying this gives:
\[ 5D = 15 \implies D = 3 \]

4. **Finding \( A \)**:
Now that we have \( D \), we can substitute it back into Equation 1 to find \( A \):
\[ A + 6(3) = 10 \]
\[ A + 18 = 10 \implies A = 10 - 18 = -8 \]

5. **Finding \( T_{20} \)**:
Now we can find \( T_{20} \):
\[ T_{20} = \frac{1}{A + (20-1)D} = \frac{1}{A + 19D} \]
Substituting the values of \( A \) and \( D \):
\[ T_{20} = \frac{1}{-8 + 19(3)} = \frac{1}{-8 + 57} = \frac{1}{49} \]

### Final Answer:
Thus, \( T_{20} = \frac{1}{49} \).