In an A.P:
given `a_(12)=37, d = 3`, find a and `S_(12)`

Answer

To solve the problem step by step, we will find the first term \( a \) and the sum of the first 12 terms \( S_{12} \) of the arithmetic progression (A.P.) given that the 12th term \( a_{12} = 37 \) and the common difference \( d = 3 \).

### Step 1: Use the formula for the nth term of an A.P.
The formula for the nth term of an A.P. is given by:
\[ a_n = a + (n - 1)d \]
For the 12th term:
\[ a_{12} = a + (12 - 1)d = a + 11d \]
Substituting the known values:
\[ 37 = a + 11 \times 3 \]

### Step 2: Simplify the equation
Calculate \( 11 \times 3 \):
\[ 11 \times 3 = 33 \]
Now, substitute this back into the equation:
\[ 37 = a + 33 \]

### Step 3: Solve for \( a \)
To find \( a \), isolate it:
\[ a = 37 - 33 \]
\[ a = 4 \]

### Step 4: Calculate the sum of the first 12 terms \( S_{12} \)
The formula for the sum of the first \( n \) terms of an A.P. is:
\[ S_n = \frac{n}{2} \times (2a + (n - 1)d) \]
Substituting \( n = 12 \), \( a = 4 \), and \( d = 3 \):
\[ S_{12} = \frac{12}{2} \times (2 \times 4 + (12 - 1) \times 3) \]

### Step 5: Simplify the sum formula
Calculate \( \frac{12}{2} \):
\[ \frac{12}{2} = 6 \]
Now calculate \( 2 \times 4 \):
\[ 2 \times 4 = 8 \]
Next, calculate \( (12 - 1) \times 3 \):
\[ 11 \times 3 = 33 \]
Now substitute these values back into the sum formula:
\[ S_{12} = 6 \times (8 + 33) \]

### Step 6: Final calculation for \( S_{12} \)
Calculate \( 8 + 33 \):
\[ 8 + 33 = 41 \]
Now calculate \( 6 \times 41 \):
\[ S_{12} = 246 \]

### Final Results
Thus, the first term \( a \) is \( 4 \) and the sum of the first 12 terms \( S_{12} \) is \( 246 \).