If C(n, 4), C(n, 5) and C(n, 6) are in AP, then what is the value of n ?

Step-by-step solution

Given that - ^n C₄, ^n C₅ and ^n C₆ are in A.P aligned & 2^n C₅= ^n C₄+ ^n C₆ & 2 n! (n-5)!5! = n! (n-4)!4! + n! (n-6)!6! aligned gathered 2 n! 5(n-5)(n-6)!4! = n! (n-4)(n-5)(n-6)!4! + n! (n-6)!6.5!4! 2 n! 5(n-5)(n-6)!4! = n! (n-6)!4! [ 1 (n-4)(n-5) + 1 6.5 ] gathered aligned & 2 (n-5)(n-6) = n^2-9 n+50 6(n-4)(n-5) & 12(n-4)=(n-5) (n^2-9 n+50 ) & n^3-15 n^2+92 n-252=0 & n^3-7 n^2-8 n^2+56 n+36 n-252=0 & n^2(n-7)-8 n(n-7)+36(n-7)=0 & (n-7) (n^2-8 n+36 )=0 & n=7 [ roots of n^2-8 n+36 is not real ] aligned