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Question

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

A
9
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B
11
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C
12
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D
14
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Solution

The correct option is D 14
nC4, nC5, nC6 are in A.P.

2× nC5= nC4+ nC6
2×n!5!×(n5)!=n!4!×(n4)!+n!6!×(n6)!

25(n5)=1(n4)(n5)+130

From options, n=14 satisfies the above equation.

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