JEE Main 2024 · MathsMultiple choiceSingle correctEasyCalculation

JEE Main 5 April 2024, Shift 1, Maths Q6

Question 6 of 90 in this shift, Maths question 6 of 30, Section A.

\frac{1}{\sqrt1+\sqrt2}+\frac{1}{\sqrt2+\sqrt3}+\ldots+\frac{1}{\sqrt{99}+\sqrt{100}}=m

and

\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\ldots+\frac{1}{99\cdot100}=n

, then the point

(m,n)

lies on the line

(1) $11x-100y=0$ Official answer
(2) $11(x-1)-100y=0$
(3) $11(x-2)-100(y-1)=0$
(4) $11(x-1)-100(y-2)=0$

Official answer

Option 1

NTA final key.

Chapter: Sequences and Series
Topic: AP and GP
Idea tested: Method of Differences for Telescoping Series

Same idea in other shifts

Asked 19× in all

5 Apr 2026, Shift 1 · Q5 $\sum_{n=1}^{10}\left(\frac{528}{n(n+1)(n+2)}\right)$ is equal to:EasySingle correct
3 Apr 2025, Shift 1 · Q7The sum $1 + 3 + 11 + 25 + 45 + 71 + \ldots$ upto 20 terms, is equal toMediumSingle correct
4 Apr 2025, Shift 2 · Q5If the sum of the first 20 terms of the series…MediumSingle correct
9 Apr 2024, Shift 2 · Q24If…HardNumerical value
30 Jan 2024, Shift 1 · Q24Let $\alpha=1^2+4^2+8^2+13^2+19^2+26^2+\ldots$ upto 10 terms and $\beta=\sum_{n=1}^{10}n^4$ . If $4\alpha-\beta=55k+40$ , then $k$ is equal…MediumNumerical value
31 Jan 2024, Shift 1 · Q5The sum of the series $\frac{1}{1-3\cdot1^2+1^4}+\frac{2}{1-3\cdot2^2+2^4}+\frac{3}{1-3\cdot3^2+3^4}+\ldots$ up to 10-terms isMediumSingle correct
6 Apr 2023, Shift 1 · Q7The sum of the first 20 terms of the series $5+11+19+29+41+\ldots$ isMediumSingle correct
8 Apr 2023, Shift 2 · Q8Let $a_n$ be the $n^{th}$ term of the series $5 + 8 + 14 + 23 + 35 + 50 + \ldots$ and $S_n = \sum_{k=1}^{n} a_k$ . Then $S_{30} - a_{40}$ …MediumSingle correct

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 108 of the 174 JEE Main shifts held from 2017 to 2026.