Arithmetic Sequence

Let's denote the common difference of the arithmetic sequence by $d$.

For the sum of an arithmetic series, we have the formula:

$$S = \frac{n}{2} \cdot (a_1 + a_n)$$

Given that $a_1 + a_3 + a_5 + a_7 + a_9 + a_{11} = -2$, we can write the sum as:

$$S_1 = \frac{6}{2} \cdot (a_1 + a_{11}) = 3 \cdot (a_1 + (a_1 + 10d)) = 3 \cdot (2a_1 + 10d)$$

$$S_1 = 6a_1 + 30d$$

Similarly, for $a_2 + a_4 + a_6 + a_8 + a_{10} + a_{12} = 1$, we get:

$$S_2 = \frac{6}{2} \cdot (a_2 + a_{12}) = 3 \cdot (a_2 + (a_2 + 11d)) = 3 \cdot (2a_2 + 11d)$$

$$S_2 = 6a_2 + 33d$$

Given that $a_1 + a_3 + a_5 + a_7 + a_9 + a_{11} = -2$ and $a_2 + a_4 + a_6 + a_8 + a_{10} + a_{12} = 1$, we can write the following system of equations:

$$6a_1 + 30d = -2$$

$$6a_2 + 33d = 1$$

Now, let's solve this system of equations. Subtracting the first equation from the second, we get:

$$6a_2 + 33d - (6a_1 + 30d) = 1 - (-2)$$

$$6a_2 - 6a_1 + 3d = 3$$

$$6(a_2 - a_1) + 3d = 3$$

$$6d + 3d = 3$$

$$9d = 3$$

$$d = \frac{1}{3}$$

Now, substitute $d = \frac{1}{3}$ into one of the original equations to find $a_1$.

Let's use the first equation:

$$6a_1 + 30 \left(\frac{1}{3}\right) = -2$$

$$6a_1 + 10 = -2$$

$$6a_1 = -12$$

$$a_1 = \boxed{-2}$$

So, $a_1 = -2$.

I got it!

Let x represent the common difference.

$a_1+a_3+a_5+a_7+a_9+a_{11} = \frac{6(a_1+a_1+10x)}{2} = 6a_1+30x = -2$

$a_2+a_4+a_6+a_8+a_{10}+a_{12} = \frac{6(a_1+x+a_1+11x)}{2} = 6a_1+36x = 1$

Solving the equation gives us $x = \frac{1}{2}, a_1 = -\frac{17}{6}$