Step 1:

The binary operation is given by

$a \ast b=\large\frac{a+b}{2}$$\;(a,b\in N)$

$\therefore b\ast a=\large\frac{b+a}{2}$ for all $a,b\in N$

Hence it is commutative.

Step 2:

$a\ast (b\ast c)=(a\ast b)\ast c$

LHS :

$a\ast \big(\large\frac{b+c}{2}\big)=\large\frac{a+\Large\frac{b+c}{2}}{2}$

$\Rightarrow \large\frac{2a+b+c}{4}$

RHS :

$(a\ast b)\ast c=\large\frac{a+b}{2}$$\ast c$

$\Rightarrow\large\frac {\Large\frac{a+b}{2}+c}{2}$

$\Rightarrow \large\frac{a+b+2c}{4}$

Clearly $LHS\neq RHS$

So it is not associative.