Let a_1,a_2,,a_2024 be an Arithmetic Progression such that a_1+(a_5+a_10+a_15++a_2020)+a_2…

Question

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 $\displaystyle\sum_{n=1}^{2024}a_n$ is equal to \_\_\_\_\_\_\_\_.

Answer: 11132

Accepted Answer

In an A.P., terms equidistant from the two ends have equal sums: $a_1+a_{2024}=a_5+a_{2020}=a_{10}+a_{2015}=\cdots$, giving $203$ equal pairs, so $203(a_1+a_{2024}) = 2233 \Rightarrow a_1+a_{2024}=11$.

$$S_{2024} = \frac{2024}{2}(a_1+a_{2024}) = 1012\times11 = 11132.$$

Solutions · 1