Given that nC\(_4\), nC\(_5\) and nC\(_6\) are the terms of a linear sequence (A.P), find...

Explanation

nC\(_4\) = \(\frac{n!}{[n-4]! 4!}\) → \(\frac{n[n-1][n-2][n-3][n-4]!}{[n-4]!4!}\)

nC\(_5\) = \(\frac{n!}{[n-5]! 5!}\) → \(\frac{n[n-1][n-2][n-3][n-4][n-5]!}{[n-5]!5!}\)

and nC\(_6\) = \(\frac{n!}{[n-6]! 6!}\) → \(\frac{n[n-1][n-2][n-3][n-4][n-5][n-6]!}{[n-6]!6!]}\)

d = U2 - U1 = U3 - U2

nC\(_5\) - nC\(_4\) = \(\frac{n(n-1)(n-2)(n-3)(n-9)}{5!}\)

nC\(_6\) - nC\(_5\) = \(\frac{n(n-1)(n-2)(n-3)(n-4)(n-11)}{5! 6}\)

nC\(_5\) - nC\(_4\) = nC\(_6\) - nC\(_5\)

\(\frac{n(n-1)(n-2)(n-3)(n-9)}{5!}\) = \(\frac{n(n-1)(n-2)(n-3)(n-4)(n-11)}{5! 6}\)

divide both sides by n(n-1)(n-2)(n-3)

\(\frac{n-9}{5!}\) = \(\frac{[n-4][n-11]}{5! 6}\)

multiply both sides by 5! * 6

6(n-9) = (n-4)(n-11)

6n - 54 = n\(^2\) -11n - 4n + 44

6n - 54 = n\(^2\) - 15n + 44

n\(^2\) - 21n + 98 = 0

n\(^2\) - 7n - 14n + 98 = 0

n(n - 7) -14(n - 7) = 0

(n-7)(n-14) = 0

n = 7 or 14

ii.

d = U2 - U1

when n = 7

d = 7C\(_5\) - 7C\(_4\)

d = \(\frac{7*6*5!}{2! * 5!}\) - \(\frac{7*6*5*4!}{3! * 4!}\)

d = 21 - 35

d = -14

when n = 14

d = 14C\(_5\) - 14C\(_4\)

d = \(\frac{14*13*12*11*10*9!}{9! * 5!}\) - \(\frac{14*13*12*11*10!}{10! * 4!}\)

d = 2002 - 1001

d = 1001