Since, nC4,nC5,nC6 are in A.P.
⇒2×nC5=nC4+nC6
⇒2×5!(n−5)!n!=4!(n−4)!n!+6!(n−6)!n!
⇒5!(n−5)!2=6!(n−4)!30+(n−5)(n−4)
⇒6×2(n−4)=30+n2−9n+20
⇒n2−21n+98=0
⇒(n−7)(n−14)=0
⇒n=7 or 14
JEE Main 2019 — Mathematics Algebra
If nC4,nC5 and nC6 are in A.P., then n can be
Held on 12 Jan 2019 · Verified 6 Jul 2026.
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