If the coefficients of (2r + 4)th and (r−2)th terms in the expansion of

Given:

$(1+x)^{18}$

We know that the coefficient of the $r$ th term in the expansion of $(1+x)^{n}$ is ${ }^{n} C_{r-1}$

Therefore, the coefficients of the $(2 r+4)$ th and $(r-2)$ th term $s$ in the given expansion are ${ }^{18} C_{2 r+4-1}$ and ${ }^{18} C_{r-2-1}$

For these coefficients to be equal, we must have

${ }^{18} C_{2 r+3}={ }^{18} C_{r-3}$

$\Rightarrow 2 r+3=r-3$ or, $2 r+3+r-3=18 \quad\left[\because{ }^{n} C_{r}={ }^{n} C_{s} \Rightarrow r=s\right.$ or $\left.r+s=n\right]$

$\Rightarrow r=-6$ or, $r=6$

Neglecting negative value We get

$r=6$