Let S = 4 , 6 , 9 and T = 9 , 10 , 11 , … , 1000 . If A = a 1 + a 2 + … + a k : k ∈ N , a…

Given,

$S=\left\{4,6,9\right\} \& T=\left\{9,10,11\dots ..1000\right\}$

Also given $A\left\{a_1+a_2+\dots ..+a_k:K\in N\right\} \& a_i\in S$

Now here by the definition of set '$A$'

$A=\left\{a:a=4x+6y+9z\right\}$

Now checking number we will get from equation $a=4x+6y+9z$

Put $x=y=0 \& z=1$ we get $a=9$

Put $x=y=1 \& z=0$ we get $a=10$

Put $x=0, y=2 \& z=0$ we get $a=12$

Put $x=1, y=0 \& z=1$ we get $a=13$

Put $x=2, y=1 \& z=o$ we get $a=14$

Put $x=0, y=1 \& z=1$ we get $a=15$

$$\begin{gathered} . \\ . \\ . \end{gathered}$$

Put $x=1, y=1 \& z=1$ we get $a=19$

Now from $20$ onwards we get all number as all number will be of type $4k, 4k+1, 4k+2 \& 4k+3$ and from equation $1, 2 \& 3$ remainder get compensated by combination of  $6y$ and $9z$,

So, except the element $11$, every element of set $T$ is of the form $4x+6y+9z$ for some $x,y,z\in W$

$\therefore T-A=\left\{11\right\}$