The coefficients of the middle terms in the binomial expansion in powers of x  of (1+ax)4  and of (1−ax)6  are same if a  equals

Given  expansion (1+ax)4 is, We have general term in the expansion (x+a)n (∴  Tr+1= nCr xn−r (a)r  be the expansion of (x+a)n) The middle term is (42+1)th  i.e. 3rd  term 3rd  term :  Tr+1= nCr xn−r (a)r  T3=T2+1= 4C2 (1)2 (ax)2 T3=T2+1=4C2 a2 x2 And in (1−ax)6,(62+1)th  i.e. 4th  term is the middle term. 4th  term :  Tr+1= nCr xn−r (a)r  And in (1−ax)6,T4=T3+1= 6C3 (1)3 (−ax)3 According to question, 4C2a2= 6C3(−a)3  (∴nCr=n!(n−r)! r!) ⇒6a2=−20a3  ⇒a=−620=−310[a≠0]