JEE Main 2022 · MathsMultiple choiceSingle correctMediumMulti-step

JEE Main 29 June 2022, Shift 2, Maths Q4

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

The sum of the infinite series

1 + \frac{5}{6} + \frac{12}{6^2} + \frac{22}{6^3} + \frac{35}{6^4} + \frac{51}{6^5} + \frac{70}{6^6} + \ldots

is equal to:

(1) $\frac{425}{216}$
(2) $\frac{429}{216}$
(3) $\frac{288}{125}$ Official answer
(4) $\frac{280}{125}$

Official answer

Option 3

NTA final key.

Chapter: Sequences and Series
Topic: Arithmetico-Geometric Series
Idea tested: Second-Order AGP Summation

Same idea in other shifts

Asked 4× in all

11 Apr 2023, Shift 2 · Q23For $k \in \mathbb{N}$ , if the sum of the series $1 + \frac{4}{k} + \frac{8}{k^2} + \frac{13}{k^3} + \frac{19}{k^4} + \ldots$ is $10$ ,…HardNumerical value
27 Jun 2022, Shift 2 · Q5Let $S=2+\frac{6}{7}+\frac{12}{7^2}+\frac{20}{7^3}+\frac{30}{7^4}+\dots..$ Then $4S$ is equal toMediumSingle correct
31 Aug 2021, Shift 2 · Q84If $\mathrm{S} = \frac{7}{5}+\frac{9}{5^2}+\frac{13}{5^3}+\frac{19}{5^4}+...$ , then $160\,\mathrm{S}$ is equal to ______________.HardNumerical 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.