In an A.P., if pth terms is `(1)/(q)` and qth term is `(1)/(p)` prove that the sum of first pq tems is

Options

A. `1/2 (pq+1)`
B. `1/3 (pq+1)`
C. `(pq+1)`
D. None

Explanation

To prove that the sum of the first \( pq \) terms of an arithmetic progression (A.P.) is \( \frac{pq + 1}{2} \), given that the \( p \)-th term is \( \frac{1}{q} \) and the \( q \)-th term is \( \frac{1}{p} \), we will follow these steps:

### Step 1: Write the formulas for the \( p \)-th and \( q \)-th terms of the A.P.
The \( n \)-th term of an A.P. can be expressed as:
\[ a_n = a + (n-1)d \]
where \( a \) is the first term and \( d \) is the common difference.

For the \( p \)-th term:
\[ a_p = a + (p-1)d = \frac{1}{q} \tag{1} \]

For the \( q \)-th term:
\[ a_q = a + (q-1)d = \frac{1}{p} \tag{2} \]

### Step 2: Set up the equations from the terms.
From equations (1) and (2), we have:
1. \( a + (p-1)d = \frac{1}{q} \)
2. \( a + (q-1)d = \frac{1}{p} \)

### Step 3: Subtract the two equations.
Subtract equation (2) from equation (1):
\[ [a + (p-1)d] - [a + (q-1)d] = \frac{1}{q} - \frac{1}{p} \]
This simplifies to:
\[ (p-1)d - (q-1)d = \frac{1}{q} - \frac{1}{p} \]
\[ (p - q)d = \frac{p - q}{pq} \]

### Step 4: Solve for \( d \).
Assuming \( p \neq q \), we can divide both sides by \( p - q \):
\[ d = \frac{1}{pq} \tag{3} \]

### Step 5: Substitute \( d \) back into one of the equations to find \( a \).
Using equation (1):
\[ a + (p-1)\left(\frac{1}{pq}\right) = \frac{1}{q} \]
\[ a + \frac{p-1}{pq} = \frac{1}{q} \]
To isolate \( a \):
\[ a = \frac{1}{q} - \frac{p-1}{pq} \]
Finding a common denominator:
\[ a = \frac{p - (p-1)}{pq} = \frac{1}{pq} \tag{4} \]

### Step 6: Find the sum of the first \( pq \) terms.
The sum of the first \( n \) terms of an A.P. is given by:
\[ S_n = \frac{n}{2} \left(2a + (n-1)d\right) \]
Substituting \( n = pq \), \( a \) from (4), and \( d \) from (3):
\[ S_{pq} = \frac{pq}{2} \left(2\left(\frac{1}{pq}\right) + (pq-1)\left(\frac{1}{pq}\right)\right) \]
\[ = \frac{pq}{2} \left(\frac{2}{pq} + \frac{pq - 1}{pq}\right) \]
\[ = \frac{pq}{2} \left(\frac{2 + pq - 1}{pq}\right) \]
\[ = \frac{pq}{2} \cdot \frac{pq + 1}{pq} \]
\[ = \frac{pq + 1}{2} \]

### Conclusion
Thus, we have proved that the sum of the first \( pq \) terms is:
\[ S_{pq} = \frac{pq + 1}{2} \]

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