# If $_{}^{n}\textrm{C}_{4}$ , $_{}^{n}\textrm{C}_{5}$ , and $_{}^{n}\textrm{C}_{6}$  are in A.P., then n can be : Option 1)$12$Option 2)$9$Option 3)$14$Option 4)$11$

Arithmetic mean of two numbers (AM) -

$A=\frac{a+b}{2}$

- wherein

It is to be noted that the sequence a, A, b, is in AP where, a and b are the two numbers.

$\frac{^{n}\textrm{C}_{4}+^{n}\textrm{C}_{6}}{2}=^{n}\textrm{C}_{5}$

$\frac{n!}{\left ( n-4 \right )!\: 4\: !}+\frac{n!}{\left ( n-6 \right )\: !\: 6!}=2\times \frac{n!}{\left ( n-5 \right )!\: 5\: !}$

$n=7,14$

Option 1)

$12$

Option 2)

$9$

Option 3)

$14$

Option 4)

$11$

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