(a) The 3rd and 19th terms of an A.P. are 13 and 77 respectively. Find the A.P.
(b) The 5th and 8th terms of an A.P. are 56 and 95 respectively. Find the 25th term of this A.P.
(c) The pth and qth terms of an A.P. are q and p respectively. Prove that its (p + q)th term will be zero.

Explanation

Let's solve the question step by step.

### Part (a)

**Given:**
- The 3rd term of an A.P. (Arithmetic Progression) is 13.
- The 19th term of the A.P. is 77.

**Let:**
- The first term of the A.P. be \( A \).
- The common difference be \( D \).

**Using the formula for the nth term of an A.P.:**
1. The 3rd term can be expressed as:
\[ A_3 = A + 2D = 13 \quad \text{(1)} \]
2. The 19th term can be expressed as:
\[ A_{19} = A + 18D = 77 \quad \text{(2)} \]

**Now, we have two equations:**
- From equation (1):
\[ A = 13 - 2D \quad \text{(3)} \]

**Substituting equation (3) into equation (2):**
\[ (13 - 2D) + 18D = 77 \]
\[ 13 + 16D = 77 \]
\[ 16D = 77 - 13 \]
\[ 16D = 64 \]
\[ D = 4 \]

**Now substitute \( D \) back into equation (3) to find \( A \):**
\[ A = 13 - 2(4) \]
\[ A = 13 - 8 = 5 \]

**Thus, the first term \( A \) is 5 and the common difference \( D \) is 4.**

**Finding the A.P.:**
The A.P. can be written as:
\[ A, A+D, A+2D, \ldots \]
Substituting the values:
\[ 5, 5+4, 5+2(4), \ldots = 5, 9, 13, 17, 21, \ldots \]

**The A.P. is:**
\[ 5, 9, 13, 17, 21, \ldots \]

### Part (b)

**Given:**
- The 5th term of the A.P. is 56.
- The 8th term of the A.P. is 95.

**Let:**
- The first term of the A.P. be \( A \).
- The common difference be \( D \).

**Using the formula for the nth term of an A.P.:**
1. The 5th term can be expressed as:
\[ A_5 = A + 4D = 56 \quad \text{(4)} \]
2. The 8th term can be expressed as:
\[ A_8 = A + 7D = 95 \quad \text{(5)} \]

**Now, we have two equations:**
- From equation (4):
\[ A = 56 - 4D \quad \text{(6)} \]

**Substituting equation (6) into equation (5):**
\[ (56 - 4D) + 7D = 95 \]
\[ 56 + 3D = 95 \]
\[ 3D = 95 - 56 \]
\[ 3D = 39 \]
\[ D = 13 \]

**Now substitute \( D \) back into equation (6) to find \( A \):**
\[ A = 56 - 4(13) \]
\[ A = 56 - 52 = 4 \]

**Thus, the first term \( A \) is 4 and the common difference \( D \) is 13.**

**Finding the 25th term:**
The 25th term can be expressed as:
\[ A_{25} = A + 24D \]
Substituting the values:
\[ A_{25} = 4 + 24(13) \]
\[ A_{25} = 4 + 312 = 316 \]

**The 25th term of this A.P. is:**
\[ 316 \]

### Part (c)

**Given:**
- The pth term of an A.P. is \( q \).
- The qth term of the A.P. is \( p \).

**Using the formula for the nth term of an A.P.:**
1. The pth term can be expressed as:
\[ A + (p-1)D = q \quad \text{(7)} \]
2. The qth term can be expressed as:
\[ A + (q-1)D = p \quad \text{(8)} \]

**From equation (7):**
\[ A = q - (p-1)D \quad \text{(9)} \]

**Substituting equation (9) into equation (8):**
\[ (q - (p-1)D) + (q-1)D = p \]
\[ q - (p-1)D + (q-1)D = p \]
\[ q + qD - pD = p + D \]
\[ q + qD - pD - p = D \]

**Rearranging gives:**
\[ D(q - p) = p - q \]

**This implies:**
\[ D = \frac{p - q}{q - p} = -1 \]

**Substituting \( D \) back into equation (9) to find \( A \):**
\[ A = q - (p-1)(-1) \]
\[ A = q + p - 1 \]

**Finding the (p + q)th term:**
The (p + q)th term can be expressed as:
\[ A + (p + q - 1)D \]
Substituting the values:
\[ = (q + p - 1) + (p + q - 1)(-1) \]
\[ = (q + p - 1) - (p + q - 1) \]
\[ = 0 \]

**Thus, the (p + q)th term is:**
\[ 0 \]

### Summary of Solutions
- **Part (a):** The A.P. is \( 5, 9, 13, 17, 21, \ldots \)
- **Part (b):** The 25th term is \( 316 \).
- **Part (c):** The (p + q)th term is \( 0 \).

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