Let `{t_(n)}` is an A.P If `t_(1) = 20 , t_(p) = q , t_(q) = p , ` find the value of m such that sum of the first m terms of the A.P is zero .

Answer

To solve the problem step by step, we will follow the reasoning laid out in the video transcript.

### Step 1: Define the terms of the A.P.
Given that \( t_1 = 20 \), we can denote the first term \( a = 20 \) and the common difference \( d \) is unknown for now.

### Step 2: Write the expressions for the terms.
The \( p \)-th term of the A.P. is given by:
\[ t_p = a + (p - 1)d = q \]
The \( q \)-th term of the A.P. is given by:
\[ t_q = a + (q - 1)d = p \]

### Step 3: Set up the equations.
From the above, we can write two equations:
1. \( 20 + (p - 1)d = q \) (Equation 1)
2. \( 20 + (q - 1)d = p \) (Equation 2)

### Step 4: Subtract Equation 1 from Equation 2.
Subtracting Equation 1 from Equation 2 gives:
\[ (20 + (q - 1)d) - (20 + (p - 1)d) = p - q \]
This simplifies to:
\[ (q - 1)d - (p - 1)d = p - q \]
\[ (q - p)d = p - q \]
Rearranging gives:
\[ (q - p)d = -(q - p) \]
Assuming \( q \neq p \), we can divide both sides by \( q - p \):
\[ d = -1 \]

### Step 5: Find the sum of the first \( m \) terms.
The sum of the first \( m \) terms of an A.P. is given by:
\[ S_m = \frac{m}{2} \left(2a + (m - 1)d\right) \]
Substituting \( a = 20 \) and \( d = -1 \):
\[ S_m = \frac{m}{2} \left(2 \cdot 20 + (m - 1)(-1)\right) \]
\[ S_m = \frac{m}{2} \left(40 - m + 1\right) \]
\[ S_m = \frac{m}{2} (41 - m) \]

### Step 6: Set the sum equal to zero.
We need to find \( m \) such that:
\[ \frac{m}{2} (41 - m) = 0 \]
This gives us two cases:
1. \( m = 0 \)
2. \( 41 - m = 0 \) which leads to \( m = 41 \)

### Conclusion:
The values of \( m \) such that the sum of the first \( m \) terms of the A.P. is zero are:
\[ m = 0 \quad \text{or} \quad m = 41 \]