JEE MAIN - Mathematics (2025 - 29th January Evening Shift - No. 25)
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
11132
Explanation
$$\mathrm{a}_1+\mathrm{a}_5+\mathrm{a}_{10}+\ldots \ldots+\mathrm{a}_{2020}+\mathrm{a}_{2024}=2233$$
In an A.P. the sum of terms equidistant from ends is equal.
$$\begin{aligned} & a_1+a_{2024}=a_5+a_{2020}=a_{10}+a_{2015} \ldots \ldots \\ & \Rightarrow 203 \text { pairs } \\ & \Rightarrow 203\left(a_1+a_{2024}\right)=2233 \end{aligned}$$
Hence,
$$\begin{aligned} & \mathrm{S}_{2024}=\frac{2024}{2}\left(\mathrm{a}_1+\mathrm{a}_{2024}\right) \\ = & 1012 \times 11 \\ = & 11132 \end{aligned}$$
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