Find the value of n

Question

If (n-1 C 4),(n-1 C 5),(n-1 C  6)are in an arithmetric progression

Answer 1

The solution is n=8 or 15. 

Let a0= n-1 C 4, and let x=n-1.

For general xCr=x!/(r!(x-r)!). When r=4 this becomes xC4=x!/(4!(x-4)!). From this it follows that a1, the next term in the AP=a0(x-4)/5 and a2=a1(x-5)/6=a0(x-4)(x-5)/30. Now we have a0, a1, a2 in terms of a0.

a1-a0=a2-a1=d, common difference.

So d=(x-4)a0/5-a0=(x-4)(x-5)a0/30-(x-4)a0/5.

(x-9)a0/5=(x-4)(x-11)a0/30.

So x-9=(x-4)(x-11)/6; 6x-54=x^2-15x+44; x^2-21x+98=0=(x-7)(x-14).

Therefore x=7 or 14 and n=x+1=8 or 15.

CHECK

For n=8 the series is 7C4, 7C5, 7C6 = 35, 21, 7 (d=-14).

For n=15 the series is 14C4, 14C5, 14C6 = 1001, 2002, 3003 (d=1001).