If the 2^ nd , 3^ rd and 4^ th terms in the expansion of (x+a)^ n are 96,216,216 respectively and n is a positive integer then a+x=

Step-by-step solution

aligned & ^n C₁ x^ n-1 a=96 ....(i) & ^n C₂ x^ n-2 a^2=216 ....(ii) & ^n C₃ x^ n-3 a^3=216 .....(iii) aligned (i) (ii) we get, aligned & ^n C₁ x ^n C₂ a = 96 216 2 n n(n-1) x a = 16 36 = 4 9 & 9 x=2(n-1) a aligned (ii) (iii) we get, aligned & ^n C₂ ^n C₃ x a =1 n(n-1) 2 x= n(n-1)(n-2) a 6 & 3 x=a(n-2) & 9 x 3 x = 2(n-1) a a(n-2) 3= 2(n-1) (n-2) & 3 n-6=2 n-2 n=4 . & 2 x a(n-1) = 4 9 & 9 x=2 a 3=6 a 3 x=2 a & ^n C₁ x^ n-1 a=96 4 x^3 3 2 x=96 & x^4=16 x=2 a=3 & a+x=5=n+1 . aligned