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 \( a_n \) of an arithmetic progression is given by:
\[ a_n = a + (n - 1) \cdot d \]
where:
- \( a \) is the first term,
- \( n \) is the term number,
- \( d \) is the common difference.

### Step 2: Substitute the known values into the formula
We know that \( n = 12 \), \( a_{12} = 37 \), and \( d = 3 \). Substituting these values into the formula gives:
\[ 37 = a + (12 - 1) \cdot 3 \]

### Step 3: Simplify the equation
Now simplify the equation:
\[ 37 = a + 11 \cdot 3 \]
\[ 37 = a + 33 \]

### Step 4: Solve for the first term \( a \)
To find \( a \), subtract 33 from both sides:
\[ a = 37 - 33 \]
\[ a = 4 \]

### Step 5: Calculate the sum of the first 12 terms \( S_{12} \)
The formula for the sum of the first \( n \) terms \( S_n \) of an A.P. is given by:
\[ S_n = \frac{n}{2} \cdot (2a + (n - 1) \cdot d) \]
For \( n = 12 \):
\[ S_{12} = \frac{12}{2} \cdot (2a + (12 - 1) \cdot d) \]

### Step 6: Substitute the known values into the sum formula
Substituting \( a = 4 \), \( d = 3 \), and \( n = 12 \) into the sum formula:
\[ S_{12} = 6 \cdot (2 \cdot 4 + 11 \cdot 3) \]

### Step 7: Simplify the expression
Calculate the values inside the parentheses:
\[ S_{12} = 6 \cdot (8 + 33) \]
\[ S_{12} = 6 \cdot 41 \]

### Step 8: Final calculation
Now multiply:
\[ S_{12} = 246 \]

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