JEE Main 2019 — Sequences And Series Question with Solution
From: JEE Main 2019 (Online) 10th April Evening Slot
Question
Let a1, a2, a3,......be an A.P. with a6 = 2. Then the common difference of this A.P., which maximises the
product a1a4a5, is :
Choose an option
Show full solutionCorrect option: C
Correct answer
C
Step-by-step explanation
first term = a, Common difference = d
a + 5d = 2
a1. a4. a5 = a(a + 3d) (a + 4d)
f(d) = (2 – 5d) (2 – 2d) (2 – d)
a + 5d = 2
a1. a4. a5 = a(a + 3d) (a + 4d)
f(d) = (2 – 5d) (2 – 2d) (2 – d)
Practice this on the real CBT interface
Solve this JEE Main question (and the rest of the Sequences And Series chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.
Solve interactively →About this question
This is a previous-year question from JEE Main 2019, 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.