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

88 Views

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

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