Q&A - Ask Doubts and Get Answers
Q

Need explanation for: - Algebra - BITSAT-2

If nC_{4}, nC_{5},nC_{6} are in A.P, then possible value of n is

  • Option 1)

    6

  • Option 2)

    12

  • Option 3)

    14

  • Option 4)

    21

 
Answers (1)
88 Views
V Vakul

As we learnt in

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.

 

2C_{5}^{n}=C_{4}^{n}\textrm{} + C_{6}^{n}\textrm{}

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

2(n-4)\times 6 = 6\times 5 + (n-4) (n-5)

12n-48=30+n^{2}-9n+20

n^{2}-21n+98=0

n^{2}-14n-7n+98=0\Rightarrow n = 14


Option 1)

6

Incorrect

Option 2)

12

Incorrect

Option 3)

14

Correct

Option 4)

21

Incorrect

Exams
Articles
Questions