If `.^(n)C_(r )=84, .^(n)C_(r-1)=36 " and" .^(n)C_(r+1)=126`, then find the value of n.

Explanation

Here, `(.^(n)C_(r))/(.^(n)C_(r-1))=(84)/(36)`
`implies(n-r+1)/(r)=(7)/(3)" "[because(.^(n)C_(r))/(.^(n)C_(r-1))=(n-r+1)/(r)]`
`implies3n-3r+3=7r` ltrgt `implies10r-3n=3` . . (i)
and `(.^(n)C_(r+1))/(.^(n)C_(r))=(n-(r+1)+1)/((r+1))=(126)/(84)" "because(.^(n)C_(r))/(.^(n)C_(r-1))=(n-r+1)/(r)]`
`implies(n-r)/(r+1)=(3)/(2)`
`implies2n-2r=3r+3`
`implies5r-2n=-3`
or `10r-4n=-6` . . . (ii)
On subtracting Eq. (ii) from Eq. (i) we get
`n=9`
From Eq. (i) we get
`10r-27=3 implies10r=30`
`thereforer=3`