Let $a_{1},a_{2},\dots,a_{2024}$ be an Arithmetic Progression such that $a_{1}+(a_{5}+a_{10}+a_{15}+\dots+a_{2020})+a_{2024}=2233$. Then $a_{1}+a_{2}+\dots+a_{2024}$ is equal to \rule{2cm}{0.4pt}.
Step-by-Step Solution
Key Concept: In an A.P., terms equidistant from the ends sum to $a_{1}+a_{n}$. The given expression pairs up to $203$ copies of $(a_{1}+a_{2024})$, giving $a_{1}+a_{2024}=11$.
In an A.P., $a_{i}+a_{n+1-i}$ is constant $=a_{1}+a_{n}$. With $n=2024$:
$$a_{5}+a_{2020}=a_{10}+a_{2015}=\dots=a_{1010}+a_{1015}=a_{1}+a_{2024}.$$
The middle terms $\{a_{5},a_{10},\dots,a_{2020}\}$ contain $\dfrac{2020-5}{5}+1=404$ entries, forming $202$ pairs. Adding $a_{1}+a_{2024}$ as the $203^{\text{rd}}$ pair:
$$203(a_{1}+a_{2024})=2233\Rightarrow a_{1}+a_{2024}=11.$$
$$\sum_{i=1}^{2024}a_{i}=\frac{2024}{2}(a_{1}+a_{2024})=1012\cdot 11=11132.$$
Correct Answer: 11132