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Friday, 9 October

JEE Main 2023 · MathsMultiple choiceSingle correctMediumMulti-step

JEE Main 8 April 2023, Shift 2, Maths Q8

Question 8 of 90 in this shift, Maths question 8 of 30, Section A.

Let ana_n be the nthn^{th} term of the series 5+8+14+23+35+50+…5 + 8 + 14 + 23 + 35 + 50 + \ldots and Sn=∑k=1nakS_n = \sum_{k=1}^{n} a_k. Then S30−a40S_{30} - a_{40} is equal to
  1. (1)11260
  2. (2)11280
  3. (3)11290Official answer
  4. (4)11310

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 8× in all
  1. 5 Apr 2026, Shift 1 · Q5∑n=110(528n(n+1)(n+2))\sum_{n=1}^{10}\left(\frac{528}{n(n+1)(n+2)}\right) is equal to:EasySingle correct
  2. 3 Apr 2025, Shift 1 · Q7The sum 1+3+11+25+45+71+…1 + 3 + 11 + 25 + 45 + 71 + \ldots upto 20 terms, is equal toMediumSingle correct
  3. 4 Apr 2025, Shift 2 · Q5If the sum of the first 20 terms of the series…MediumSingle correct
  4. 9 Apr 2024, Shift 2 · Q24If…HardNumerical value
  5. 31 Jan 2024, Shift 1 · Q5The sum of the series 11−3⋅12+14+21−3⋅22+24+31−3⋅32+34+…\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. 6 Apr 2023, Shift 1 · Q7The sum of the first 20 terms of the series 5+11+19+29+41+…5+11+19+29+41+\ldots isMediumSingle correct
  7. 18 Mar 2021, Shift 1 · Q68132−1+152−1+172−1+…+1(201)2−1\frac{1}{3^2-1}+\frac{1}{5^2-1}+\frac{1}{7^2-1}+\ldots+\frac{1}{(201)^2-1} is equal to :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 42 of the 174 JEE Main shifts held from 2017 to 2026.