If the 7th term of a H.P. is 8 and the 8th term is 7, then its 15th term is

Options

A. 16
B. 14
C. 27/14
D. 56/15

Answer

To solve the problem, we need to find the 15th term of a Harmonic Progression (H.P.) given that the 7th term is 8 and the 8th term is 7.

### Step-by-Step Solution:

1. **Understanding the nth term of H.P.**:
The nth term of a Harmonic Progression can be expressed in terms of the first term \( A \) and the common difference \( D \) of the corresponding Arithmetic Progression (A.P.). The formula for the nth term of H.P. is:
\[ T_n = \frac{1}{A + (n-1)D} \]

2. **Setting up the equations**:
For the 7th term \( T_7 \):
\[ T_7 = \frac{1}{A + 6D} = 8 \]
This can be rearranged to:
\[ A + 6D = \frac{1}{8} \quad \text{(Equation 1)} \]

For the 8th term \( T_8 \):
\[ T_8 = \frac{1}{A + 7D} = 7 \]
This can be rearranged to:
\[ A + 7D = \frac{1}{7} \quad \text{(Equation 2)} \]

3. **Subtracting the equations**:
Now we will subtract Equation 1 from Equation 2:
\[ (A + 7D) - (A + 6D) = \frac{1}{7} - \frac{1}{8} \]
Simplifying this gives:
\[ D = \frac{1}{7} - \frac{1}{8} \]
To subtract these fractions, we need a common denominator, which is 56:
\[ D = \frac{8}{56} - \frac{7}{56} = \frac{1}{56} \]

4. **Finding the value of A**:
Now that we have \( D \), we can substitute it back into Equation 1 to find \( A \):
\[ A + 6 \left(\frac{1}{56}\right) = \frac{1}{8} \]
This simplifies to:
\[ A + \frac{6}{56} = \frac{1}{8} \]
Converting \( \frac{1}{8} \) to a fraction with a denominator of 56:
\[ \frac{1}{8} = \frac{7}{56} \]
Thus:
\[ A + \frac{6}{56} = \frac{7}{56} \]
Solving for \( A \):
\[ A = \frac{7}{56} - \frac{6}{56} = \frac{1}{56} \]

5. **Finding the 15th term**:
Now we can find the 15th term \( T_{15} \):
\[ T_{15} = \frac{1}{A + 14D} \]
Substituting the values of \( A \) and \( D \):
\[ T_{15} = \frac{1}{\frac{1}{56} + 14 \left(\frac{1}{56}\right)} = \frac{1}{\frac{1 + 14}{56}} = \frac{1}{\frac{15}{56}} = \frac{56}{15} \]

### Final Answer:
The 15th term of the H.P. is \( \frac{56}{15} \).