Let 1+x22(1+x)n=∑r=1n+4 arxr. If a1, a2, a3 are in A.P., then n is equal to

1+2x2+x41+nC1x+nC2x2+nC3x3+…=a0+a1x+a2x2+a3x3+…⇒a1=nC1,a2=nC2+2,a3=nC3+2 nC1Now, a1, a2, a3 are in A.P. implies2 nC2+2=nC1+nC3+2 nC1⇒n(n−1)+4=3n+16n(n−1)(n−2)⇒6n2−n+24=18n+n3−3n2+2n⇒n3−9n2+26n−24=0⇒(n−2)(n−3)(n−4)=0⇒n=2,3,4