Sequences & Series
AP Sum via Symmetric Pairing
nta_pyq_2025_apr
Grade 11
Question:
Let $a_1, a_2, \ldots, a_{2024}$ be an AP. If $a_1+(a_5+a_{10}+a_{15}+\cdots+a_{2020})+a_{2024}=2233$, then $\displaystyle\sum_{i=1}^{2024}a_i$ equals
Step-by-Step Solution
Key Concept: Group $a_1+a_{2024}$ with each pair $(a_{5k}+a_{2025-5k})$ — all equal to $a_1+a_{2024}$ by AP symmetry — giving 203 equal sums.
The sequence $a_5, a_{10},\ldots,a_{2020}$ has $\frac{2020-5}{5}+1=404$ terms.
By AP symmetry, $a_k+a_{2025-k}=a_1+a_{2024}$ for all $k$. Pairing:
$(a_1+a_{2024})+(a_5+a_{2020})+(a_{10}+a_{2015})+\cdots+(a_{1010}+a_{1015})$
$= 1+202 = 203$ pairs, each summing to $a_1+a_{2024}$.
$203(a_1+a_{2024})=2233\Rightarrow a_1+a_{2024}=11$.
$\sum_{i=1}^{2024}a_i=\frac{2024}{2}(a_1+a_{2024})=1012\times11=11132$.
Correct Answer: 11132