​T3​=nC2​xn−6a2...........(i)T4​=nC3​xn−9a3.......(ii)​ T5​=nC4​xn−12a4﻿...........(iii) After dividing (i) and (ii), we get n−23​×ax3​=812​........(iv) n−34​×ax3​=38​......(v) After solving equation (iv) and (v), we get n=6-3 now general term of expansion is Tt+1​=6Cr​x6−3rar 6−3r=0⇒r=2 T3​ is term independent of x.