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If nC4, nC5 and nC6 are in A.P., then n can be:
- 11
- 12
- 9
- 14
Correct answer: 14
Solution
The condition that nC4, nC5, and nC6 are in arithmetic progression implies that twice the middle term (nC5) equals the sum of the other two terms (nC4 + nC6). This relationship leads to a specific value for n, and through calculations, it can be determined that n must be 14 for this equality to hold.
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