Let `T_(r)` be the rth term of an AP, for r=1,2,… If for some positive integers m and n, we have `T_(m)=(1)/(n) and T_(n)=(1)/(m)," the "T_(m+n)` equals

Options

A. `(1)/(mn)`
B. `(1)/(m)+(1)/(n)`
C. `(1)/(m)`
D. 0

Answer

To solve the problem, we need to find the value of \( T_{m+n} \) given the conditions \( T_m = \frac{1}{n} \) and \( T_n = \frac{1}{m} \).

### Step-by-Step Solution:

1. **Understanding the Terms of an AP**:
The r-th term of an arithmetic progression (AP) can be expressed as:
\[ T_r = a + (r-1)d \]
where \( a \) is the first term and \( d \) is the common difference.

2. **Setting Up the Equations**:
From the problem, we have:
\[ T_m = a + (m-1)d = \frac{1}{n} \quad \text{(1)} \]
\[ T_n = a + (n-1)d = \frac{1}{m} \quad \text{(2)} \]

3. **Subtracting the Equations**:
To eliminate \( a \), we can subtract equation (2) from equation (1):
\[ (a + (m-1)d) - (a + (n-1)d) = \frac{1}{n} - \frac{1}{m} \]
This simplifies to:
\[ (m-n)d = \frac{1}{n} - \frac{1}{m} \]

4. **Finding a Common Denominator**:
The right side can be rewritten with a common denominator:
\[ \frac{1}{n} - \frac{1}{m} = \frac{m - n}{mn} \]
So we have:
\[ (m-n)d = \frac{m-n}{mn} \]

5. **Dividing by \( (m-n) \)**:
Assuming \( m \neq n \), we can divide both sides by \( (m-n) \):
\[ d = \frac{1}{mn} \]

6. **Substituting \( d \) Back**:
Now we can substitute \( d \) back into either equation (1) or (2) to find \( a \). Let's use equation (1):
\[ a + (m-1)\left(\frac{1}{mn}\right) = \frac{1}{n} \]
Rearranging gives:
\[ a = \frac{1}{n} - \frac{m-1}{mn} \]
Simplifying this:
\[ a = \frac{m - 1}{mn} - \frac{m-1}{mn} = \frac{1}{mn} \]

7. **Finding \( T_{m+n} \)**:
Now we can find \( T_{m+n} \):
\[ T_{m+n} = a + (m+n-1)d \]
Substituting \( a \) and \( d \):
\[ T_{m+n} = \frac{1}{mn} + (m+n-1)\left(\frac{1}{mn}\right) \]
This simplifies to:
\[ T_{m+n} = \frac{1}{mn} + \frac{m+n-1}{mn} = \frac{1 + m + n - 1}{mn} = \frac{m+n}{mn} \]

### Final Answer:
\[ T_{m+n} = \frac{m+n}{mn} \]