In the expansion of (1+x)18, we have Tp+1= 18Cpxp
and T(r−2)=T(r−3)+1= 18C(r−3).xr−3⋯(2)
It is given that the coefficients of T(2r+4) and T(r−2) are equal.
So, from (1) and (2), we get
⇒(2r+3)=(r−3) or (2r+3)+(r−3)=18 [∵ nCp=nCq⇒p=q or p+q=n]
Now, (2r+3)=(r−3) or (2r+3)+(r−3)=18
⇒r=−6 or r=6
[∵ the negative value of r is not permissible].