JEE Main 2018MathematicsSequences and SeriesHardMCQ

JEE Main 2018 — Sequences and Series Question with Solution

JEE Main 2018 (16 Apr Online)

Question

Let 1x1, 1x2,…,1xn(xi≠0 for i=1, 2,…., n) be in A.P. such that x1=4 and x21=20 . If n is the least positive integer for which xn>50, then ∑i=1n1xi is equal to

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Show full solutionCorrect option: C
Correct answer
C134

Step-by-step explanation

Given 1x1,1x2,………,1xn are in A.P. and x1=4 and x21=20.

⇒a=14

Given, x21=20.

⇒14+20·d=120

⇒d=−1100

Given, xn>50.

⇒1xn<150

⇒14- n-1100< 150⇒n>24

∵ n=25

Now, ∑i=1251xi=2522×14-1100×24=134

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About this question

This is a previous-year question from JEE Main 2018, covering the Sequences and Series chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.