In a H.P., `p^(th)` term is q and `q^(th)` term is p then `pq^(th)` term is

Options

A. p
B. 1
C. pq
D. q

Answer

To solve the problem, we need to find the `pq^(th)` term in a Harmonic Progression (H.P.) given that the `p^(th)` term is `q` and the `q^(th)` term is `p`.

### Step-by-Step Solution:

1. **Understanding H.P. and A.P. Relationship**:
In a Harmonic Progression, the reciprocals of the terms form an Arithmetic Progression (A.P.).
Let the `p^(th)` term in H.P. be `T_p = q`, which implies that the `p^(th)` term in A.P. is `1/q`.
Similarly, let the `q^(th)` term in H.P. be `T_q = p`, which implies that the `q^(th)` term in A.P. is `1/p`.

2. **Formulating the A.P. Terms**:
Let the first term of the A.P. be `a` and the common difference be `d`.
The `p^(th)` term in A.P. is given by:
\[ T_p = a + (p - 1)d = \frac{1}{q} \quad \text{(1)} \]
The `q^(th)` term in A.P. is given by:
\[ T_q = a + (q - 1)d = \frac{1}{p} \quad \text{(2)} \]

3. **Setting Up the Equations**:
From equations (1) and (2), we have:
\[ a + (p - 1)d = \frac{1}{q} \quad \text{(1)} \]
\[ a + (q - 1)d = \frac{1}{p} \quad \text{(2)} \]

4. **Subtracting the Equations**:
Subtract equation (2) from equation (1):
\[ (a + (p - 1)d) - (a + (q - 1)d) = \frac{1}{q} - \frac{1}{p} \]
Simplifying gives:
\[ (p - q)d = \frac{1}{q} - \frac{1}{p} \]
\[ (p - q)d = \frac{p - q}{pq} \]
If \( p \neq q \), we can divide both sides by \( p - q \):
\[ d = \frac{1}{pq} \]

5. **Finding the First Term `a`**:
Substitute \( d = \frac{1}{pq} \) back into equation (1):
\[ a + (p - 1)\left(\frac{1}{pq}\right) = \frac{1}{q} \]
Rearranging gives:
\[ a = \frac{1}{q} - \frac{p - 1}{pq} \]
\[ a = \frac{p - 1 - p + 1}{pq} = \frac{0}{pq} = 0 \]

6. **Finding the `pq^(th)` Term**:
The `pq^(th)` term in A.P. is given by:
\[ T_{pq} = a + (pq - 1)d \]
Substituting \( a = 0 \) and \( d = \frac{1}{pq} \):
\[ T_{pq} = 0 + (pq - 1)\left(\frac{1}{pq}\right) \]
\[ T_{pq} = \frac{pq - 1}{pq} = 1 - \frac{1}{pq} \]

7. **Finding the `pq^(th)` Term in H.P.**:
Since we need the `pq^(th)` term in H.P.:
\[ T_{pq}^{H.P.} = \frac{1}{T_{pq}^{A.P.}} = \frac{1}{1 - \frac{1}{pq}} = \frac{pq}{pq - 1} \]

### Final Answer:
The `pq^(th)` term in H.P. is \( \frac{pq}{pq - 1} \).