The second, third and fourth terms in the binomial expansion `(x+a)^n`are 240, 720 and 1080, respectively. Find x, a and n.

Explanation

By the binomial expansion , we have
`(x+a)^(n) = .^(n) C _(0) x ^(n)+ .^(n) C _ (1) x^(n-1) a+ .^(n)C_(2) x ^(n-2) a^(2) + . ^(n) C _(3) x ^(n-3) a^(3) + ...`
`rArr (x+a)^(n) = x^(n) + nx ^(n-1) a + 1/2 n ( n-1) x^((n-2)) a^(2)`
`+ 1/6 n ( n-1) ( n-2) x^(n-3 a^(3) + ...`
` :. T_(2) = 240 rArr nx^(n-1) a = 240, " "` ...(i)
`T_(3) = 720 rArr n(n-1) x^(n-2) a^(2) = 1440," "` ...(ii)
`T_(4) = 1080 rArr n ( n-1) (n-2) x^(n-3) a^(3) = 6480. " "` ...(iii)
On dividing (ii) by (i) , we get
`((n-1)a)/x = 1440/240 rArr a/x = 6/((n-1)). " "` ...(iv)
On dividing (iii) by (ii), we get
`((n-2)a)/x = 6480/1440 rArr a/ x = 9/(2(n-2))." "` ...(v)
Equating the values of `a/x` from (iv) and (v), we get
`6/(n-1) = 9 /2(n-2) rArr 12(n-2) = 9 ( n-1) rArr 3n = 15 rArr n=5.`
Putting n=5 in (i) we get `x^(4) a = 48" "` ...(vi)
Putting n=5 (iv) , we get `a/x = 6/4 rArr a/x = 3/2 rArr a = 3/2 x.`
Putting `a=3/2 x ` in (vi), we get
`x^(4) 3/2 x = 48 rArr x^(5) = 48 xx 2/3 = 32 = 2^(5) rArr x= 2 .`
Hence, ` x=2, a=3, and n=5.`