The problem is to find the first term $a_1$ and the common difference $d$ of an arithmetic sequence $\{a_n\}$ given the sum of the first 8 terms $S_8 = 48$ and the sum of the first 12 terms $S_{12} = 168$.

1. Problem Description

The problem is to find the first term $a_1$ and the common difference $d$ of an arithmetic sequence $\{a_n\}$ given the sum of the first 8 terms $S_8 = 48$ and the sum of the first 12 terms $S_{12} = 168$.

2. Solution Steps

The sum of the first $n$ terms of an arithmetic sequence is given by:

$S_n = \frac{n}{2}[2a_1 + (n-1)d]$

Using the given information, we have:

$S_8 = \frac{8}{2}[2a_1 + (8-1)d] = 48$

$4(2a_1 + 7d) = 48$

$2a_1 + 7d = 12$ (Equation 1)

$S_{12} = \frac{12}{2}[2a_1 + (12-1)d] = 168$

$6(2a_1 + 11d) = 168$

$2a_1 + 11d = 28$ (Equation 2)

Subtracting Equation 1 from Equation 2, we get:

$(2a_1 + 11d) - (2a_1 + 7d) = 28 - 12$

$4d = 16$

$d = 4$

Substituting $d = 4$ into Equation 1:

$2a_1 + 7(4) = 12$

$2a_1 + 28 = 12$

$2a_1 = -16$

$a_1 = -8$

3. Final Answer

$a_1 = -8$

$d = 4$