JEE Main 2025 · MathsMultiple choiceSingle correctMediumMulti-step

JEE Main 2 April 2025, Shift 1, Maths Q19

Question 19 of 75 in this shift, Maths question 19 of 25, Section A.

Let

a_1, a_2, a_3,\ldots

be in an A.P. such that

\sum_{k=1}^{12}a_{2k-1}=-\frac{72}{5}a_1

a_1\neq0

. If

\sum_{k=1}^{n}a_k=0

, then n is :

(1)10
(2)11Official answer
(3)17
(4)18

Official answer

Option 2

NTA final key.

Chapter: Sequences and Series
Topic: AP and GP
Idea tested: AP Sum of n terms

Same idea in other shifts

Asked 15× in all

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7 Apr 2025, Shift 2 · Q6Let $a_n$ be the $n^{\mathrm{th}}$ term of an A.P. If $S_n=a_1+a_2+a_3+\ldots+a_n=700$ , $a_6=7$ and $S_7=7$ , then $a_n$ is equal to :MediumSingle correct
6 Apr 2024, Shift 2 · Q7A software company sets up $m$ number of computer systems to finish an assignment in 17 days. If 4 computer systems crashed on the start…MediumSingle correct
8 Apr 2024, Shift 1 · Q25Let the positive integers be written in the form : 1 2 3 4 5 6 7 8 9 10 ... If the $k^{\mathrm{th}}$ row contains exactly $k$ numbers for…MediumNumerical valueHas a figure
9 Apr 2024, Shift 1 · Q26If a function $f$ satisfies $f(m+n)=f(m)+f(n)$ for all $m,n\in\mathbf{N}$ and $f(1)=1$ , then the largest natural number $\lambda$ such…MediumNumerical value

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 42 of the 174 JEE Main shifts held from 2017 to 2026.