Given $C_{n - 3,} C_{n - 2,} C_{n - 1}$ are in A.P., then

n = 14

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n = 7

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n = 12

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C_{1}, C_{2}, C_{3}

are in H.P

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Solution

The correct option is B $n = 7$
Since $_{n - 3}^{n} C_{n - 2},^{n} C_{n - 1}^{n} C$ are in AP, it means that $_{3}^{n} C,_{2}^{n} C,_{1}^{n} C$ are in AP
Thus, $_{3}^{n} C +_{1}^{n} C = {2.}_{2}^{n} C \Rightarrow \frac{n (n - 1) (n - 2)}{6} + n = 2. \frac{n (n - 1)}{2} \Rightarrow n^{2} - 3 n + 2 + 6 = 6 n - 6 \Rightarrow n^{2} - 9 n + 14 = 0 \Rightarrow (n - 2) (n - 7) = 0 => n = 2, 7$
Since $n > 2 \Rightarrow n = 7$

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