Question

If the coefficients of r r +1 r+2terms in(1+x)^14 are in ap then r

Answer 1

Use the formula, the general term in the expansion of $(a+b)^n$ is

$T_{r+1}=C(r,n)a^rb^{n-r},r=0,1,2,...,n$.

Using this formula, the $r,r+1,r+2$ terms in the expansion of $(1+x)^{14}$ are

$T_{r}=C(r-1,14)x^{15-r}\\ T_{r+1}=C(r,14)x^{14-r}\\ T_{r+2}=C(r+1,14)x^{13-r}$

The coefficients $C(r-1,14),C(r,14),C(r+1,14)$ are in AP. That is

$2C(r,14)=C(r+1,14)+C(r-1,14)\\ 2\frac{14!}{(14-r)!r!} =\frac{14!}{(13-r)!(r+1)!} +\frac{14!}{(15-r)!(r-1)!} \\ 2(15-r)(r+1)=(15-r)(14-r)+r(r+1)\\ 2(-r^2+14r+15)=210-29r+r^2+r^2+r\\ 4r^2-56r+180=0\\ r^2-14r+45=0\\ (r-9)(r-5)=0\\ r=5,9$

The possible values if $r$ are $r=5,9$