JEE Main 2018MathematicsSequences and SeriesMediumMCQ

JEE Main 2018 — Sequences and Series Question with Solution

JEE Main 2018 (15 Apr)

Question

If x1, x2, ….., xn and 1h1, 1h2, ….., 1hn are two A.P.s such that x3=h2=8 & x8=h7=20, then x5·h10 is equal to

Choose an option

Show full solutionCorrect option: D
Correct answer
D2560

Step-by-step explanation

Let d1 be the common difference of the A.P. x1,x2,…, then x8-x3=5 d1=12⇒

⇒x5=x3+2 d1=8+2×125=12.8

Let d2 be the common difference of the other sequence then

5 d2=120-18=-340⇒d2

⇒1 h10=1 h7+3 d2=1200⇒h10=200

⇒x5·h10=12.8×200=2560

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 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.