Answer :

Given that nC4, nC5 and nC6 are in A.P.


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.


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