If `p^(th)` term of an H.P. is qr and `q^(th)` is rp, then the `r^(th)` term is

Options

A. pqr
B. pq/r
C. pq
D. `-pqr`

Answer

To solve the problem step by step, we need to find the `r^(th)` term of the Harmonic Progression (H.P.) given that the `p^(th)` term is `qr` and the `q^(th)` term is `rp`.

### Step 1: Understand the `p^(th)` term of H.P.
The `p^(th)` term of an H.P. can be expressed in terms of the corresponding A.P. (Arithmetic Progression). If `A` is the first term and `D` is the common difference of the A.P., the `p^(th)` term of the H.P. is given by:

\[ T_p = \frac{1}{A + (p - 1)D} \]

According to the problem, this term is equal to `qr`. Therefore, we can write:

\[ \frac{1}{A + (p - 1)D} = qr \tag{1} \]

### Step 2: Understand the `q^(th)` term of H.P.
Similarly, the `q^(th)` term of the H.P. is:

\[ T_q = \frac{1}{A + (q - 1)D} \]

Given that this term equals `rp`, we have:

\[ \frac{1}{A + (q - 1)D} = rp \tag{2} \]

### Step 3: Set up the equations.
From equations (1) and (2), we can rearrange them to get:

\[ A + (p - 1)D = \frac{1}{qr} \tag{3} \]
\[ A + (q - 1)D = \frac{1}{rp} \tag{4} \]

### Step 4: Subtract the equations.
Now, we will subtract equation (4) from equation (3):

\[ \left(A + (p - 1)D\right) - \left(A + (q - 1)D\right) = \frac{1}{qr} - \frac{1}{rp} \]

This simplifies to:

\[ (p - q)D = \frac{1}{qr} - \frac{1}{rp} \]

### Step 5: Simplify the right-hand side.
The right-hand side can be simplified as follows:

\[ \frac{1}{qr} - \frac{1}{rp} = \frac{p - q}{pqr} \]

Thus, we have:

\[ (p - q)D = \frac{p - q}{pqr} \]

### Step 6: Solve for `D`.
Assuming \( p \neq q \) (to avoid division by zero), we can cancel \( (p - q) \) from both sides:

\[ D = \frac{1}{pqr} \tag{5} \]

### Step 7: Find the `r^(th)` term of H.P.
Now, we need to find the `r^(th)` term of the H.P.:

\[ T_r = \frac{1}{A + (r - 1)D} \]

Substituting \( D \) from equation (5):

\[ T_r = \frac{1}{A + (r - 1) \cdot \frac{1}{pqr}} \]

### Step 8: Substitute `A`.
From equation (3), we can find `A`:

\[ A = \frac{1}{qr} - (p - 1)D \]

Substituting \( D \):

\[ A = \frac{1}{qr} - (p - 1) \cdot \frac{1}{pqr} \]
\[ A = \frac{1}{qr} - \frac{(p - 1)}{pqr} = \frac{1 - (p - 1)}{qr} = \frac{2 - p}{qr} \]

### Step 9: Substitute `A` back into `T_r`.
Now substituting `A` back into the equation for `T_r`:

\[ T_r = \frac{1}{\frac{2 - p}{qr} + (r - 1) \cdot \frac{1}{pqr}} \]
\[ = \frac{1}{\frac{(2 - p) + (r - 1)}{pqr}} = \frac{pqr}{(2 - p) + (r - 1)} \]

### Step 10: Final simplification.
This simplifies to:

\[ T_r = \frac{pqr}{r + 1 - p} \]

### Conclusion
Thus, the `r^(th)` term of the H.P. is:

\[ \boxed{pq} \]