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Question

If nC4, nC5 and nC6 are in A.P., then n can be:

A
14
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B
11
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C
9
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D
12
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Solution

The correct option is A 14
If a,b,c are in A.P, then 2b=a+c

So applying the same condition, we have

2×(nC5)=(nC6)+(nC4)

2×n!5!(n5)!=n!6!(n6)!+n!4!(n4)!

25!(n5)!=16!(n6)!+14!(n4)!

25!(n5)(n6)!=16!(n6)!+14!(n4)(n5)(n6)!

25!(n5)=16!+14!(n4)(n5)

25×4!×(n5)=16×5×4!+14!(n4)(n5)

25×(n5)=16×5+1(n4)(n5)

25×(n5)1(n4)(n5)=16×5

12n78=n29n+20

n221n+98=0

(n7)(n14)=0

n=14 or n=7

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