If the coeffecients of rth,(r+1)th and (r+2)th terms in the expansion of (1+x)14 are in A.P., then r is/are
Given : rth,(r+1)th and (r+2)th terms in the expansion of (1+x)14 are in A.P.
⇒mCr−1,mCr,mCr+1 are in A.P.
⇒2 mCr=mCr−1+mCr+1
⇒2=mCr−1+mCr+1mCr
⇒rm−r+1+m−rr+1⇒m2−m(4r+1)+4r2−2=0
Substitute m=14 ...... [Given]
⇒r=5,9
Hence, options A and D are correct.