If the second, third and fourth terms in the expansion of (x+a)n  are 240,720 and 1080 respectively, then the value of n  is

We know that rth  term in the expansion of (x+a)n is Tr+1=nCr xn−r ar Given second term   T2=240    nC1 xn−1 a=240   ​(∴ nC1=n,r=1)   ⇒n​ xn−1 a=240                         …(1) Given third termT3=720   nC2 xn−2 a2=720 (∴ nC2=n(n−1)2,r=2) ⇒n(n−1)2  xn−2 a2=720            …(2) Given fourth termT4=1080 nC3 xn−3 a3=1080 (∴ nC3=n (n−1) (n−2)6,r=3)  ⇒n (n−1) (n−2)6 xn−3 a3=1080   ….(3)Squaring Eq (2) and dividing it by product of Eq (1) and Eq (3), we get (n (n−1)2)26n2(n−1) (n−2)=720×720240×1080 ⇒  6 (n−1)4 (n−2)=2  ⇒  6n−6=  8n−16  ∴  n=5