If the coefficient of   kth  , (k+1)th,(k+2)nd  terms in the expansion of (1+x)n are in A..P then

Given expansion is(1+x)n  then we have general term of  (x+a)n is Tr+1=nCr(x)n−r⋅(a)r Here kth  term : T(k−1)+1=nCk−1(1)n−k+1⋅(x)k−1=nCk−1(x)k−1 ∴ co-efficient of kth term is nCk−1 (k+1)th  term :Tk+1=nCk(1)n−k⋅(x)k=nCk⋅xk ∴ co-efficient of (k+1)th term is nCk (k+2)nd  term :T(k+1)+1=nCk+1(1)n−k−1⋅(x)k+1=nCk+1⋅(x)k+1 co-efficient of (k+2)nd term is nCk+1 According to problem Given co-efficient of terms  nCk−1 ,nCk ,nCk+1 are in A.P⇒2nCk=nCk−1+nCk+1 (∴ if a,b,c are in A.P then 2b=a+c )⇒2=nCk−1nCk+nCk+1nCk(∴nCrnCr−1=n−r+1r) ⇒2=kn−k+1+n−kk+1 ⇒2=k(k+1)+(n−k)(n−k+1)(k+1)(n−k+1) ⇒2(k+1)(n−k+1)=k(k+1)+(n−k)(n−k+1) ⇒2[nk−k2+k+n−k+1]=[k2+k+n2−nk+n−nk+k2−k] ⇒2nk−2k2+2n+2−2k2+2nk−n2−n=0 ⇒−4k2+4nk+n−n2+2=0 ⇒n+2=n2−4nk+4k2 ⇒n+2=(n−2k)2