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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

Answers (1)

best_answer

 

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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