JEE Main 2018 — Sequences And Series Question with Solution
From: JEE Main 2018 (Online) 16th April Morning Slot
Question
Let (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 then is equal to :
Choose an option
Show full solutionCorrect option: D
Correct answer
D
Step-by-step explanation
are in A.P.
x1 = 4 and x21 = 20
Let 'd' be the common difference of this A.P.
its 21st term =
d = d =
Also xn > 50(given).
xn =
1 + (n 1) ( ) 4 <
(n 1) <
n > 23 n > 24
Therefore n = 25.
= =
x1 = 4 and x21 = 20
Let 'd' be the common difference of this A.P.
its 21st term =
d = d =
Also xn > 50(given).
xn =
1 + (n 1) ( ) 4 <
(n 1) <
n > 23 n > 24
Therefore n = 25.
= =
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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.