If the mth term of an AP is 1/n and nth term is 1/m, then find the sum to mn terms.

Options

A. ( mn – 1)/4
B. (mn + 1)/4
C. (mn + 1)/2
D. (mn - 1)/2

Answer

To solve the problem step by step, we will use the properties of arithmetic progressions (AP).

### Step 1: Understand the Given Information
We are given that the m-th term of an AP is \( \frac{1}{n} \) and the n-th term is \( \frac{1}{m} \).

### Step 2: Write the General Formula for the m-th and n-th Terms
The m-th term of an AP can be expressed as:
\[ a_m = a + (m-1)d \]
where \( a \) is the first term and \( d \) is the common difference.

Similarly, the n-th term can be expressed as:
\[ a_n = a + (n-1)d \]

### Step 3: Set Up the Equations
From the information given:
1. \( a + (m-1)d = \frac{1}{n} \) (Equation 1)
2. \( a + (n-1)d = \frac{1}{m} \) (Equation 2)

### Step 4: Subtract the Two Equations
Subtract Equation 2 from Equation 1:
\[ (a + (m-1)d) - (a + (n-1)d) = \frac{1}{n} - \frac{1}{m} \]
This simplifies to:
\[ (m-n)d = \frac{1}{n} - \frac{1}{m} \]

### Step 5: Simplify the Right Side
The right side can be simplified:
\[ \frac{1}{n} - \frac{1}{m} = \frac{m-n}{mn} \]
Thus, we have:
\[ (m-n)d = \frac{m-n}{mn} \]

### Step 6: Solve for d
Assuming \( m \neq n \), we can divide both sides by \( m-n \):
\[ d = \frac{1}{mn} \]

### Step 7: Substitute d Back to Find a
Now, substitute \( d \) back into Equation 1 to find \( a \):
\[ a + (m-1)\left(\frac{1}{mn}\right) = \frac{1}{n} \]
This simplifies to:
\[ a + \frac{m-1}{mn} = \frac{1}{n} \]
Rearranging gives:
\[ a = \frac{1}{n} - \frac{m-1}{mn} \]
Finding a common denominator:
\[ a = \frac{m - (m-1)}{mn} = \frac{1}{mn} \]

### Step 8: Find the Sum of mn Terms
The sum of the first \( N \) terms of an AP is given by:
\[ S_N = \frac{N}{2} \left(2a + (N-1)d\right) \]
For \( N = mn \):
\[ S_{mn} = \frac{mn}{2} \left(2\left(\frac{1}{mn}\right) + (mn-1)\left(\frac{1}{mn}\right)\right) \]
This simplifies to:
\[ S_{mn} = \frac{mn}{2} \left(\frac{2}{mn} + \frac{mn - 1}{mn}\right) \]
\[ = \frac{mn}{2} \left(\frac{2 + mn - 1}{mn}\right) = \frac{mn}{2} \cdot \frac{mn + 1}{mn} \]
\[ = \frac{mn + 1}{2} \]

### Final Answer
Thus, the sum of the first \( mn \) terms is:
\[ \boxed{\frac{mn + 1}{2}} \]