Mathematics · Sequence and Series

JEE Main 2025 — 29 January, Evening Shift — Question 64

Let $a_1, a_2, \ldots, a_{2024}$ be an arithmetic progression such that $a_1 + (a_5 + a_{10} + a_{15} + \cdots + a_{2020}) + a_{2024} = 2233.$ Then $a_1 + a_2 + a_3 + \cdots + a_{2024}$ is equal to:

Answer: 11132

Numerical answer — enter this value.

Step-by-step solution

${a_1} + {a_{5\;}} + {a_{10}} + \ldots . + {a_{2020}} + {a_{2024}} = 2233$

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

${a_1} + {a_{2024}} = {a_5} + {a_{2020}} = {a_{10}} + {a_{2015}}= \ldots ..$ =203 pairs

= 203 ( ${a_1} + {a_{2024}}) = 2233$

Hence, ${S_{2024 =}}\frac{{2024}}{2}\left( {{a_1} + {a_{2024}}} \right)$ = $1012×11 =11132$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 29 January, Evening Shift
Subject: Mathematics
Chapter: Sequence and Series
Topic: Arithmetic Progression