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JEE Main 202529 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
29 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Sequences and Series

JEE Main 2025 · 29 January · Evening Shift · Mathematics · Sequences and Series

Let a_1, a_2, , a_ 2024 be an Arithmetic Progression such that...

Let a1,a2,…,a2024a_1, a_2, \ldots, a_{2024} be an Arithmetic Progression such that a1+(a5+a10+a15+…+a2020)+a2024=2233a_1+\left(a_5+a_{10}+a_{15}+\ldots+a_{2020}\right)+a_{2024}=2233. Then a1+a2+a3+…+a2024a_1+a_2+a_3+\ldots+a_{2024} is equal to _______

Correct answer

0

Step-by-step solution

As a1+a5+a10+…+a2020+a2024=2233a_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) a1+a2024=a5+a2020=a10+a2015=…⏟203 pairs \underbrace{a_1+a_{2024}=a_5+a_{2020}=a_{10}+a_{2015}=\ldots}_{203 \text{ pairs }}⇒203(a1+a2024)=2233⇒a1+a2024=11\begin{aligned} & \Rightarrow \quad 203\left(a_1+a_{2024}\right)=2233 \\ & \Rightarrow \quad a_1+a_{2024}=11\end{aligned}NowNow\begin{aligned} \sum_{i=1}^{2024} a_i & =S_{2024}=\frac{2024}{2}\left[a_1+a_{2024}\right] \\ & =1012(11) \\ & =11132\end{aligned}$$
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