The first term, common difference and last term of an A.P. are 12,6 and 252 respectively, Find the sum of all terms of this A.P.

Answer

To find the sum of all terms of the given Arithmetic Progression (A.P.) where the first term \( a = 12 \), the common difference \( d = 6 \), and the last term \( l = 252 \), we will follow these steps:

### Step 1: Find the number of terms \( n \)
The formula for the \( n \)-th term of an A.P. is given by:
\[ a_n = a + (n - 1) \cdot d \]
Here, \( a_n = l = 252 \), \( a = 12 \), and \( d = 6 \). Plugging in these values, we get:
\[ 252 = 12 + (n - 1) \cdot 6 \]

### Step 2: Simplify the equation
Subtract \( 12 \) from both sides:
\[ 252 - 12 = (n - 1) \cdot 6 \]
\[ 240 = (n - 1) \cdot 6 \]

### Step 3: Divide both sides by 6
\[ n - 1 = \frac{240}{6} \]
\[ n - 1 = 40 \]

### Step 4: Solve for \( n \)
Adding \( 1 \) to both sides gives:
\[ n = 40 + 1 = 41 \]

### Step 5: Find the sum of the first \( n \) terms \( S_n \)
The formula for the sum of the first \( n \) terms of an A.P. is:
\[ S_n = \frac{n}{2} \cdot (2a + (n - 1) \cdot d) \]
Substituting \( n = 41 \), \( a = 12 \), and \( d = 6 \):
\[ S_{41} = \frac{41}{2} \cdot (2 \cdot 12 + (41 - 1) \cdot 6) \]

### Step 6: Calculate \( S_{41} \)
First, calculate \( 2 \cdot 12 \):
\[ 2 \cdot 12 = 24 \]
Next, calculate \( (41 - 1) \cdot 6 \):
\[ 40 \cdot 6 = 240 \]
Now, substitute these values back into the sum formula:
\[ S_{41} = \frac{41}{2} \cdot (24 + 240) \]
\[ S_{41} = \frac{41}{2} \cdot 264 \]

### Step 7: Final calculation
Calculating \( \frac{41 \cdot 264}{2} \):
\[ S_{41} = 41 \cdot 132 = 5412 \]

Thus, the sum of all terms of the A.P. is \( \boxed{5412} \).