If the coefficients of the r th, (r+1) th, (r−2) th terms in the expansion of (1+x)14are in AP, then the the sum of all possible values of r is ____.

According to the question, 14Cr−1,14Cr,14Cr+1 are in A.P.,so b=a+c2⇒ 214Cr=14Cr−1+14Cr+1or   2×14!(14−r)!r!=14!(14−r+1)!(r−1)!+14!(14−r−1)!(r+1)!or   2(14−r)(13−r)!r(r−1)!=1(15−r)(14−r)(13−r)!(r−1)!+1(13−r)!(r+1)r(r−1)!or   2(14−r)r=1(15−r)(14−r)+1r(r+1)or   2(14−r)r−1r(r+1)=1(15−r)(14−r)or  3r−12r(r+1)=1(15−r) r=5 or 9