If the `7^(th)` terms of a H.P. is 8 and the `8^(th)` term is 7. Then find the `28^(th)` term

Answer

To solve the problem step by step, we need to find the 28th term of a Harmonic Progression (H.P.) given that the 7th term is 8 and the 8th term is 7.

### Step 1: Understand the relationship between H.P. and A.P.
If \( A, B, C \) are in H.P., then \( \frac{1}{A}, \frac{1}{B}, \frac{1}{C} \) are in Arithmetic Progression (A.P.).

### Step 2: Set up the terms in A.P.
Given:
- The 7th term of H.P. is \( 8 \) (let's denote it as \( H_7 = 8 \)).
- The 8th term of H.P. is \( 7 \) (let's denote it as \( H_8 = 7 \)).

The corresponding terms in A.P. will be:
- \( A_7 = \frac{1}{H_7} = \frac{1}{8} \)
- \( A_8 = \frac{1}{H_8} = \frac{1}{7} \)

### Step 3: Write the equations for the A.P.
The general term of an A.P. can be expressed as:
- \( A_n = a + (n-1)d \)

For the 7th term:
\[ A_7 = a + 6d = \frac{1}{8} \]
For the 8th term:
\[ A_8 = a + 7d = \frac{1}{7} \]

### Step 4: Set up the system of equations
We have the following two equations:
1. \( a + 6d = \frac{1}{8} \) (Equation 1)
2. \( a + 7d = \frac{1}{7} \) (Equation 2)

### Step 5: Subtract the equations to find \( d \)
Subtract Equation 1 from Equation 2:
\[ (a + 7d) - (a + 6d) = \frac{1}{7} - \frac{1}{8} \]
This simplifies to:
\[ d = \frac{1}{7} - \frac{1}{8} \]

### Step 6: Calculate \( d \)
Finding a common denominator (56):
\[ d = \frac{8}{56} - \frac{7}{56} = \frac{1}{56} \]

### Step 7: Substitute \( d \) back to find \( a \)
Now substitute \( d \) back into Equation 1:
\[ a + 6 \left(\frac{1}{56}\right) = \frac{1}{8} \]
This simplifies to:
\[ a + \frac{6}{56} = \frac{1}{8} \]
Converting \( \frac{1}{8} \) to a fraction with a denominator of 56:
\[ \frac{1}{8} = \frac{7}{56} \]
Now we have:
\[ a + \frac{3}{28} = \frac{7}{56} \]
Subtract \( \frac{6}{56} \) from both sides:
\[ a = \frac{7}{56} - \frac{6}{56} = \frac{1}{56} \]

### Step 8: Find the 28th term in A.P.
The 28th term \( A_{28} \) is given by:
\[ A_{28} = a + 27d \]
Substituting the values of \( a \) and \( d \):
\[ A_{28} = \frac{1}{56} + 27 \left(\frac{1}{56}\right) = \frac{1 + 27}{56} = \frac{28}{56} = \frac{1}{2} \]

### Step 9: Convert back to H.P.
The 28th term of the H.P. is the reciprocal of the 28th term of the A.P.:
\[ H_{28} = \frac{1}{A_{28}} = \frac{1}{\frac{1}{2}} = 2 \]

### Final Answer:
The 28th term of the H.P. is \( 2 \).