To solve the series \( 1 - 5 + 9 - 13 + \ldots + 41 - 45 \), let’s analyze the pattern.

1. The series consists of alternating positive and negative numbers.
2. The positive terms are: \( 1, 9, 17, 25, 33, 41 \)
3. The negative terms are: \( 5, 13, 21, 29, 37, 45 \)

**Positive Terms Calculation:**
The positive terms form an arithmetic sequence:
- First term \( a_1 = 1 \)
- Common difference \( d = 8 \)

The last term is \( 41 \). We can find the number of terms \( n \) using the formula for the \( n \)-th term of an arithmetic sequence:
\[
a_n = a_1 + (n-1)d
\]
Setting \( a_n = 41 \):
\[
41 = 1 + (n-1) \cdot 8
\]
\[
40 = (n-1) \cdot 8
\]
\[
n-1 = 5 \implies n = 6
\]

Now, to find the sum of positive terms:
\[
S_{\text{positive}} = \frac{n}{2} (a_1 + a_n) = \frac{6}{2} (1 + 41) = 3 \cdot 42 = 126
\]

**Negative Terms Calculation:**
The negative terms also form an arithmetic sequence:
- First term \( b_1 = 5 \)
- Common difference \( d = 8 \)

The last term is \( 45 \). Using the same approach:
\[
b_n = b_1 + (n-1)d
\]
Setting \( b_n = 45 \):
\[
45 = 5 + (n-1) \cdot 8
\]
\[
40 = (n-1) \cdot 8
\]
\[
n-1 = 5 \implies n = 6
\]

Now, to find the sum of negative terms:
\[
S_{\text{negative}} = \frac{n}{2} (b_1 + b_n) = \frac{6}{2} (5 + 45) = 3 \cdot 50 = 150
\]

**Final Calculation:**
Now we can combine the sums:
\[
\text{Total Sum} = S_{\text{positive}} - S_{\text{negative}} = 126 - 150 = -24
\]

Thus, the total sum of the series \( 1 - 5 + 9 - 13 + \ldots + 41 - 45 \) is \(-24\).