JEE Main 2019MathematicsSequences and SeriesEasyMCQ

JEE Main 2019 — Sequences and Series Question with Solution

JEE Main 2019 (09 Jan Shift 1)

Question

If a, b and c be three distinct real numbers in G.P. and a+b+c=xb, then x cannot be:

Choose an option

Show full solutionCorrect option: B
Correct answer
B2

Step-by-step explanation

Given a, b, c are in G.P.

Let common ratio of G.P. be r , then b=ar, c=ar2

Given equation can be written as

a+ar+ar2=x. a. r

⇒x=r+1r+1

As we know r>0, then r+1r≥2⇒x≥3

r<0, then r+1r≤-2⇒x≤-1

⇒x∈-∞,-1∪3, ∞

Hence, x ≠2.

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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.