Find the sum of first 25 terms of an A.P. whose 4th term is 13 and 22nd term is 67.

Answer

To find the sum of the first 25 terms of an arithmetic progression (A.P.) where the 4th term is 13 and the 22nd term is 67, we can follow these steps:

### Step 1: Identify the formulas
The \( n \)-th term of an A.P. can be expressed as:
\[ T_n = a + (n-1)d \]
where \( a \) is the first term and \( d \) is the common difference. The sum of the first \( n \) terms \( S_n \) is given by:
\[ S_n = \frac{n}{2} \times (2a + (n-1)d) \]

### Step 2: Set up the equations using the given terms
From the problem, we know:
- The 4th term \( T_4 = 13 \):
\[ T_4 = a + (4-1)d = a + 3d = 13 \quad \text{(Equation 1)} \]
- The 22nd term \( T_{22} = 67 \):
\[ T_{22} = a + (22-1)d = a + 21d = 67 \quad \text{(Equation 2)} \]

### Step 3: Solve the equations
We have the two equations:
1. \( a + 3d = 13 \)
2. \( a + 21d = 67 \)

Subtract Equation 1 from Equation 2:
\[ (a + 21d) - (a + 3d) = 67 - 13 \]
This simplifies to:
\[ 18d = 54 \]
So, we find:
\[ d = \frac{54}{18} = 3 \]

### Step 4: Substitute \( d \) back to find \( a \)
Now substitute \( d = 3 \) back into Equation 1:
\[ a + 3(3) = 13 \]
This simplifies to:
\[ a + 9 = 13 \]
Thus:
\[ a = 13 - 9 = 4 \]

### Step 5: Calculate the sum of the first 25 terms
Now that we have \( a = 4 \) and \( d = 3 \), we can find \( S_{25} \):
\[ S_{25} = \frac{25}{2} \times (2a + (25-1)d) \]
Substituting the values of \( a \) and \( d \):
\[ S_{25} = \frac{25}{2} \times (2 \times 4 + 24 \times 3) \]
Calculating inside the parentheses:
\[ 2 \times 4 = 8 \]
\[ 24 \times 3 = 72 \]
Thus:
\[ S_{25} = \frac{25}{2} \times (8 + 72) = \frac{25}{2} \times 80 \]
Calculating further:
\[ S_{25} = \frac{25 \times 80}{2} = \frac{2000}{2} = 1000 \]

### Final Answer
The sum of the first 25 terms of the A.P. is \( \boxed{1000} \).