Let $$a_1, a_2, \ldots, a_{2024}$$ be an Arithmetic Progression such that $$a_1 + (a_5 + a_{10} + a_{15} + \cdots + a_{2020}) + a_{2024} = 2233$$. Then $$a_1 + a_2 + a_3 + \cdots + a_{2024}$$ is equal to _______

Given an AP $$a_1, a_2, \ldots, a_{2024}$$ such that:

$$a_1 + (a_5 + a_{10} + a_{15} + \cdots + a_{2020}) + a_{2024} = 2233$$

The terms $$a_5, a_{10}, a_{15}, \ldots, a_{2020}$$ form an AP with first term $$a_5$$, common difference $$5d$$, and the number of terms: from 5 to 2020 with step 5, that's $$\frac{2020-5}{5} + 1 = 404$$ terms.

For an AP: $$a_n = a_1 + (n-1)d$$.

The sum of the subsequence terms:

$$a_5 + a_{10} + \cdots + a_{2020} = \frac{404}{2}(a_5 + a_{2020}) = 202(a_5 + a_{2020})$$

In an AP: $$a_5 + a_{2020} = a_1 + a_{2024}$$ (since $$5 + 2020 = 2025 = 1 + 2024$$).

So the given equation becomes:

$$a_1 + a_{2024} + 202(a_1 + a_{2024}) = 2233$$

$$203(a_1 + a_{2024}) = 2233$$

$$a_1 + a_{2024} = 11$$

The sum of the full AP:

$$S_{2024} = \frac{2024}{2}(a_1 + a_{2024}) = 1012 \times 11 = 11132$$

The answer is 11132.