Let $a_1, a_2, \ldots, a_{2024}$ be an Arithmetic Progression such that $a_1+\left(a_5+a_{10}+a_{15}+\ldots+a_{2020}\right)+a_{2024}=2233$. Then $a_1+a_2+a_3+\ldots+a_{2024}$ is equal to _________.
Answer (integer)
11132
Solution
<p>$$\mathrm{a}_1+\mathrm{a}_5+\mathrm{a}_{10}+\ldots \ldots+\mathrm{a}_{2020}+\mathrm{a}_{2024}=2233$$</p>
<p>In an A.P. the sum of terms equidistant from ends is equal.</p>
<p>$$\begin{aligned}
& a_1+a_{2024}=a_5+a_{2020}=a_{10}+a_{2015} \ldots \ldots \\
& \Rightarrow 203 \text { pairs } \\
& \Rightarrow 203\left(a_1+a_{2024}\right)=2233
\end{aligned}$$</p>
<p>Hence,</p>
<p>$$\begin{aligned}
& \mathrm{S}_{2024}=\frac{2024}{2}\left(\mathrm{a}_1+\mathrm{a}_{2024}\right) \\
= & 1012 \times 11 \\
= & 11132
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Sequences and Series · Topic: Arithmetic Progression
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