JEE Main 2022MathematicsSequences and SeriesMediumMCQ

JEE Main 2022 — Sequences and Series Question with Solution

JEE Main 2022 (29 Jul Shift 1)

Question

If 120-a40-a+140-a60-a+……+ 1180-a200-a=1256, then the maximum value of a is

Choose an option

Show full solutionCorrect option: C
Correct answer
C212

Step-by-step explanation

Given,120-a40-a+140-a60-a+……+ ...+ 1180-a200-a=1256

So, on simplifying we get,

⇒120120-a-140-a+140-a-160-a.....

+…+1180-a-1200-a=1256

⇒120120-a-1200-a=1256

⇒20-a200-a=256×9

⇒a2-220a+1696=0

⇒a=8,212

So, maximum value of a is 212.

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About this question

This is a previous-year question from JEE Main 2022, 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.