The 8th and 14th term of a H.P. are 1/2 and 1/3, respectively. Find its 20th term. Also, find its general term.

Explanation

To solve the problem, we need to find the 20th term and the general term of the given Harmonic Progression (H.P.) where the 8th term is \( \frac{1}{2} \) and the 14th term is \( \frac{1}{3} \).

### Step 1: Understanding the Harmonic Progression
In a Harmonic Progression, the reciprocals of the terms form an Arithmetic Progression (A.P.). Let's denote the first term of the A.P. as \( a \) and the common difference as \( d \).

The terms of the H.P. can be expressed as:
- 1st term: \( \frac{1}{a} \)
- 2nd term: \( \frac{1}{a + d} \)
- 3rd term: \( \frac{1}{a + 2d} \)
- n-th term: \( \frac{1}{a + (n-1)d} \)

### Step 2: Setting up the equations
From the problem, we know:
- The 8th term of the H.P. is \( \frac{1}{2} \):
\[ \frac{1}{a + 7d} = \frac{1}{2} \implies a + 7d = 2 \quad \text{(Equation 1)} \]

- The 14th term of the H.P. is \( \frac{1}{3} \):
\[ \frac{1}{a + 13d} = \frac{1}{3} \implies a + 13d = 3 \quad \text{(Equation 2)} \]

### Step 3: Solving the equations
Now we have two equations:
1. \( a + 7d = 2 \)
2. \( a + 13d = 3 \)

We can subtract Equation 1 from Equation 2:
\[ (a + 13d) - (a + 7d) = 3 - 2 \]
This simplifies to:
\[ 6d = 1 \implies d = \frac{1}{6} \]

### Step 4: Finding \( a \)
Now, substitute \( d \) back into Equation 1 to find \( a \):
\[ a + 7 \left(\frac{1}{6}\right) = 2 \]
\[ a + \frac{7}{6} = 2 \]
Subtract \( \frac{7}{6} \) from both sides:
\[ a = 2 - \frac{7}{6} = \frac{12}{6} - \frac{7}{6} = \frac{5}{6} \]

### Step 5: Finding the 20th term
Now we can find the 20th term of the H.P.:
\[ \text{20th term} = \frac{1}{a + 19d} \]
Substituting the values of \( a \) and \( d \):
\[ = \frac{1}{\frac{5}{6} + 19 \times \frac{1}{6}} = \frac{1}{\frac{5}{6} + \frac{19}{6}} = \frac{1}{\frac{24}{6}} = \frac{1}{4} \]

### Step 6: Finding the general term
The general term of the H.P. is given by:
\[ \text{General term} = \frac{1}{a + (n-1)d} \]
Substituting the values of \( a \) and \( d \):
\[ = \frac{1}{\frac{5}{6} + (n-1) \times \frac{1}{6}} = \frac{1}{\frac{5 + (n-1)}{6}} = \frac{6}{5 + n - 1} = \frac{6}{n + 4} \]

### Final Answers
- The 20th term of the H.P. is \( \frac{1}{4} \).
- The general term of the H.P. is \( \frac{6}{n + 4} \).