If the coefficients of 5th, 6th and 7th terms in the expansion of (1+x)nare in A.P., then n =

Coefficient of T5 is nC4,that of T6 is nC5 and that of T7 is nC6. According to the condition, 2nC5=nC4+nC6.Hence,2n!(n−5)!5!=n!(n−4)!4!+n!(n−6)!6!or 21(n−5)5=1(n−4)(n−5)+16×5After solving, we get n = 7 or 14.