If `""^(n)C_(4),^(n)C_(5)" and "^(n)C_(6)` are in A.P. then find n.

Options

A. 9
B. 11
C. 14
D. 12

Answer

C

Explanation

If `""^(n)C_(4), ""^(n)C_(5)` and `""^(n)C_(6)` are in AP, then
`2*""^(n)C_(5) = ""^(n)C_(4) + ""^(n)C_(6)` [If a, b, c are in AP, then 2b = a + c]
`rArr 2(n!)/(5!(n - 5)!) = (n!)/(4!(n - 4)!) + (n!)/(6!(n - 6)!)" "[because ""^(n)C_(r) = (n!)/(r!(n - r)!)]`
`rArr (2)/(5*4!(n - 5)(n - 6)!)`
`=(1)/(4!(n-4)(n-5)(n - 6)!)+(1)/(6*5*4!(n - 6)!)`
`rArr (2)/(5(n-5)) = (1)/((n-4)(n-5))+(1)/(30)`
`rArr (2)/(5(n-5)) = (30+ (n - 4)(n - 5))/(30(n - 4)(n - 5))`
`rArr 12(n - 4) = 30 + n^(2) - 9n + 20`
`rArr` `n^(2) - 21n + 98 = 0`
`rArr` `n^(2) - 14n - 7n + 98 = 0`
`rArr n(n - 14) - 7(n - 14) = 0`
`rArr` `(n - 7)(n - 14) = 0`
`rArr` `n = 7` or 14