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 _______

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Aicharya · Accepted Answer

As $a_1+a_5+a_{10}+\ldots+a_{2020}+a_{2024}=2233$ ...(1)
We know in arithmetic progression.
Sum of terms equidistant from ends is equal
$\therefore$ from (1)
$\underbrace{a_1+a_{2024}=a_5+a_{2020}=a_{10}+a_{2015}=\ldots}_{203 \text { pairs }}$
$\begin{aligned} & \Rightarrow \quad 203\left(a_1+a_{2024}\right)=2233 \\ & \Rightarrow \quad a_1+a_{2024}=11\end{aligned}$
Now $\begin{aligned} \sum_{i=1}^{2024} a_i & =S_{2024}=\frac{2024}{2}\left[a_1+a_{2024}\right] \\ & =1012(11) \\ & =11132\end{aligned}$

Let $a_1, a_2, \ldots, a_{2024}$ be an Arithmetic Progression such that…

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