If the 2nd, 3rd and 4th terms in the expansion of `(x+a)^n` are 240, 720 and 1080 respectively, find `x ,\ a ,\ ndot`

Explanation

Given , ` T_(2) = T_(1+1) = ""^(n)C_(1) . x^(n-1) y = 240 ` …(i)
` T_(3) = T_( 2 + 1) = ""^(n)C_(2) . x^(x - 2) y ^(2) = 720 ` …(ii)
and ` T_(4) = T_(3 + 1) = ""^(n)C_(2) * x^(n-3) * y^(4) = 1080 ` …(iii)
ON dividing Eq (ii) Eq (i) we get
` (""^(n)C_(2) * x^(n-2) * y^(2))/(""^(n)C_(1) * x^(n-1) * y) = (720)/(24)`
` rArr ((n-2+ 1)/(2)) * (y)/(x) = 3 rArr (y)/(x) = (6)/(n-1)` ...(iv)
Also , dividing Eq (iii) by Eq. (ii) , we get
`(""^(n)C_(3) * x^(n-3) * y^(3))/(""^(n)C_(2) * x^(n-2) * y^(2)) = (1080)/(720 )`
` rArr ((n-3 +1)/(3)) * (y)/(x) = (3)/(2) rArr (y)/(x) = (9)/(2(n-2))` (v)
From Eqs (iv) and (v) , we get
`(6)/(n-1) = (9)/(2(n-2))`
` rArr 12n - 24 = 9 n - 9 `
` rArr 3n = 15 `
` therefore n = 5 `
From Eq(iv), we get ` y = (3)/(2) x `
From Eq (i) and (vi) , we get
`""^(5)C_(1) x^(4) y = 240 rArr 5 * x^(4) (3)/(2) x = 240 `
` therefore x^(5) = 32 = 2^(5) rArr x = 2 `
From Eq (vi) , we get y = 3
Hence , x = 2 y = 3 and n = 5