If the coefficients of 5th , 6th  and 7th  terms in the expansion of (1+x)n,n∈N , are in A.P., then n  is equal to

The coefficients of xr−1,xr,xr+1  in the expansion of (1+x)n  are  nCr−1, nCr, nCr+1  Respectively. Since these are in A.P. we have (suppose a,b,c are in AP  then 2b=a+c ) here a= nCr−1, b= nCr, c= nCr+1  then   2nCr= 2Cr−1+ nCr+1  wkt  nCr=  nr n−r ⇒  2 nr n−r=nr−1 n−r+1+nr+1 n−r−1  ⇒  2 nr r−1(n−r) n−r−1=nr−1 (n−r+1) (n−r) n−r−1+nr (r+1) r−1 n−r−1  (Both sides cancellation of terms n ,r−1 ,n−r−1  then we get) ⇒  2r (n−r)=  1(n−r+1) (n−r)+1(r+1) r ⇒  2r (n−r)=  1(n−r+1) (n−r)+1(r+1) r⇒2r (n−r)=(r+1) r+(n−r+1) (n−r)(n−r+1) (n−r)(r+1) r  (Both sides cancellation of terms,r(n−r)  then we get)  ⇒21=(r+1) r+(n−r+1) (n−r)(n−r+1) (r+1)  Now cross multiplication we get    2(n−r+1) (r+1)=r (r+1)+(n−r+1) (n−r) ⇒2 (n−r+1) (r+1)=r (r+1)+(n−r+1) (n−r)⇒2 (nr+n−r2−r+r+1)=r2+r+n2−nr−rn+r2+n−r ⇒  n2−n (4r+1)+4 r2−2=0 Deduction. If the coefficients of 5th ,6th  and 7th  terms in the expansion of (1+x)n  are in A.P. then  nC4,nC5,nC6 ⇒  2 nC5= nC4+nC6   ⇒  n2−(4.5+1)n  +4(25)−2=0⇒ n2−21 n+98  =0  ⇒  n=7,  14   Solving the above Eqn  we get  7,14