Since the coefficient of (r+1)th term in the expansion of (1+x)n=nCr ∴ In the expansion of (1+x)18
coefficient of (2r+4)th term =18C2r+3,
Similarly, coefficient of (r−2)th term in the expansion of (1+x)18=18Cr−3
If nCr=nCs then r+s=n
So, 18C2r+3=18Cr−3 gives 2r+3+r−3=18 ⇒3r=18⇒r=6.