If the coefficients of rth, (r + 1)th and (r +2)th terms in the expansion of (1 + x)14 are in AP, then r is /are

Coefficient of rth, (r + 1) th and (r +2)th terms in (1 +x)14 are  14Cr−1,14Cr,14Cr+1 respectivelyNow, according to the question, 2(14Cr)=14Cr−1+14Cr+1On dividing both sides by  14Cr, we get2= 14Cr−1 14Cr+ 14Cr+1 14Cr ⇒ 2=r14−r+1+14−(r+1)+1r+1 ⇒ 2=r15−r+14−rr+1 ⇒ 2(15−r)(r+1)=r(r+1)+(15−r)(14−r) ⇒ −2r2+28r+30=2r2−28r+210 ⇒ 4r2−56r+180=0⇒r2−14r+45=0 ⇒ (r−9)(r−5)=0 ⇒ r=5,9