JEE Main 2022 · MathsMultiple choiceSingle correctMediumMulti-step

JEE Main 29 July 2022, Shift 1, Maths Q4

Question 4 of 90 in this shift, Maths question 4 of 30, Section A.

\frac{1}{(20 - a)(40 - a)} + \frac{1}{(40 - a)(60 - a)} + \ldots + \frac{1}{(180 - a)(200 - a)} = \frac{1}{256}

, then the maximum value of a is :

(1)198
(2)202
(3)212Official answer
(4)218

Official answer

Option 3

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
5 Apr 2024, Shift 1 · Q6If $\frac{1}{\sqrt1+\sqrt2}+\frac{1}{\sqrt2+\sqrt3}+\ldots+\frac{1}{\sqrt{99}+\sqrt{100}}=m$ and…EasySingle 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

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.