JEE Main 2020 · MathsMultiple choiceSingle correctMediumMulti-step

JEE Main 5 September 2020, Shift 1, Maths Q69

Question 69 of 75 in this shift, Maths question 19 of 25.

If S is the sum of the first 10 terms of the series

\tan^{-1}\left(\frac{1}{3}\right)+\tan^{-1}\left(\frac{1}{7}\right)+\tan^{-1}\left(\frac{1}{13}\right)+\tan^{-1}\left(\frac{1}{21}\right)+....,

then tan(S) is equal to :

(1) $\frac{5}{6}$ Official answer
(2) $\frac{10}{11}$
(3) $-\frac{6}{5}$
(4) $\frac{5}{11}$

Official answer

Option 1

NTA final key (Sep 2020).

Chapter: Inverse Trigonometric Functions
Topic: Inverse Trig Identities
Idea tested: Telescoping Inverse Trig Series

Same idea in other shifts

Asked 5× in all

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28 Jun 2022, Shift 2 · Q18The value of $\lim_{n\to\infty} 6\tan\left\{\sum_{r=1}^{n}\tan^{-1}\left(\frac{1}{r^2+3r+3}\right)\right\}$ is equal to :MediumSingle correct
26 Aug 2021, Shift 2 · Q77If $\sum_{\mathrm{r}=1}^{50}\tan^{-1}\frac{1}{2\mathrm{r}^2}=\mathrm{p}$ , then the value of $\tan\mathrm{p}$ is :MediumSingle correct
24 Feb 2021, Shift 1 · Q90 $\lim_{n \to \infty} \tan\left\{\sum_{r=1}^{n} \tan^{-1}\left(\frac{1}{1 + r + r^2}\right)\right\}$ is equal to ___________.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 108 of the 174 JEE Main shifts held from 2017 to 2026.