Given that nC4 4 , nC5 5 and nC6 6 are the terms of a linear sequence (A.P), find the : i. value of n ii. common differences of the sequence.

Answer Details

nC4 ${}_4$ = n![n−4]!4! $\frac{\mathrm{<span\; style="font-weight:\; var(--bs-body-font-weight);"> \to </span>}}{}$n[n−1][n−2][n−3][n−4]![n−4]!4! $\frac{\mathrm{<br>}}{}$

nC5 ${}_5$ = n![n−5]!5! $\frac{\mathrm{<span\; style="font-weight:\; var(--bs-body-font-weight);"> \to </span>}}{}$n[n−1][n−2][n−3][n−4][n−5]![n−5]!5!]

and nC6 ${}_6$ = n![n−6]!6! $\frac{\mathrm{<br>}}{}$  → n[n−1][n−2][n−3][n−4][n−5][n−6]![n−6]!6!] $\frac{\mathrm{<br>}}{}$

d = U2 - U1 = U3 - U2

nC5 ${}_5$ - nC4 ${}_4$ = n(n−1)(n−2)(n−3)(n−9)5! $\frac{\mathrm{<br>}}{}$

nC6 ${}_6$ - nC5 ${}_5$ = n(n−1)(n−2)(n−3)(n−4)(n−11)5!6 $\frac{\mathrm{<br>}}{}$

nC5 ${}_5$ - nC4 ${}_4$ = nC6 ${}_6$ - nC5 ${}_5$

n(n−1)(n−2)(n−3)(n−9)5! $\frac{\mathrm{<br>}}{}$ = n(n−1)(n−2)(n−3)(n−4)(n−11)5!6 $\frac{\mathrm{<br>}}{}$

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

n−95!
 = [n−4][n−11]5!6 $\frac{<br>}{}$

multiply both sides by 5! * 6

6(n-9) = (n-4)(n-11)
6n - 54 = n2 ${}^2$ -11n - 4n + 44
6n - 54 = n2 ${}^2$ - 15n + 44
n2 ${}^2$ - 21n + 98 = 0
n2 ${}^2$ - 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 = 7C5 ${}_5$ - 7C4 ${}_4$

d = 7∗6∗5!2!∗5! $\frac{7\ast 6\ast 5!}{2!\ast 5!}$ - 7∗6∗5∗4!3!∗4! $\frac{7\ast 6\ast 5\ast 4!}{3!\ast 4!}$

d = 21 - 35
d = -14

when n = 14

d = 14C5 ${}_5$ - 14C4 ${}_4$

d = 14∗13∗12∗11∗10∗9!9!∗5! $\frac{\mathrm{<br>}}{}$ - 14∗13∗12∗11∗10!10!∗4! $\frac{\mathrm{<br>}}{}$

d = 2002 - 1001
d = 1001