Question

If nc4, nc5 and nc6 are in

a.p., then the value of n is

Answer 1

It is given that C(n,4) and C(n,5) and C(n,6) are in A.P. where C(n,r)=$\frac{n!}{(n-r)!\times r!}$

The three numbers a,b,c are in A.P then , → 2 b = a+c

→ 2 C(n,5) = C(n,4) + C(n,6)

→C(n,5) -C(n,4) =C(n,6) - C(n,5)

→$\frac{n!}{(n-5)!\times5!}-\frac{n!}{(n-4)!\times4!}=\frac{n!}{(n-6)!\times6!}-\frac{n!}{(n-5)!\times5!}$

Cancelling n! from numerator, we get

and (n-5)!=(n-5)(n-6)! and (n-4)!=(n-4)(n-5)(n-6)! and 6!=6×5×4 and 5!=5×4!

Also ,Cancelling (n-6)! and 4! from denominator

→$\frac{1}{5\times(n-5)}-\frac{1}{(n-5)(n-4)}=\frac{1}{6\times5}-\frac{1}{5\times(n-5)}$

→ $\frac{1}{n-5}[\frac{1}{5}-\frac{1}{n-4}]=\frac{1}{5}[\frac{1}{6}-\frac{1}{n-5}]$

→n²-21 n + 98 =0

→(n-14)(n-7)=0

n=14  or n=7

Answer 2

Answer:

Step-by-step explanation:

We know that for three numbers a, b, c are in A.P, the following condition holds,

⇒ 2b = a + c

So,

⇒ 2nC5 = nC4 + nC6

Adding 2nC5 on both sides we get,

⇒ 4nC5 = nC4 + nC5 + nC5 + nC6

We know that nCr + nCr + 1 = n + 1Cr + 1

⇒ 4nC5 = n + 1C5 + n + 1C6

⇒ 4nC5 = n + 2C6

We know that

And also n! = n(n – 1)(n – 2)…………2.1

⇒ 24(n – 4) = n2 + 2n + n + 2

⇒ 24n – 96 = n2 + 3n + 2

⇒ n2 – 21n + 98 = 0

⇒ n2 – 14n – 7n + 98 = 0

⇒ n(n – 14) – 7(n – 14) = 0

⇒ (n – 7)(n – 14) = 0

⇒ n – 7 = 0 or n – 14 = 0

⇒ n = 7 or n = 14

∴ The values of n are 7 and 14.

Hope it helps :)