Verified Solution • 2025–26 Session moderate

QJEEM25Mad0a1: JEE Main 2025 Mathematics - Class 11 JEEM25Mad0a1 Solution: Let be an Arithmetic Progression such that . Then is equal to…

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 _________.

Step-by-Step Verified Solution

a1+a5+a10+……+a2020+a2024=2233

In an A.P. the sum of terms equidistant from ends is equal.

a1+a2024=a5+a2020=a10+a2015……⇒203 pairs ⇒203(a1+a2024)=2233

Hence,

S2024=20242(a1+a2024)=1012×11=11132

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