If coefficient of 9th, 10th and 11th terms of (1 + x)n are in A.P., then value of n can be

2 nC9=nC8+nC10⇒2=9n−8+n−910⇒20(n−8)=90+(n−8)(n−9)⇒n2−37n+322=0⇒(n−14)(n−23)=0Thus, n can be 23