Let S = \ 4, 6, 9\ and T = \ 9, 10, 11, , 1000\ . If A = \ a _ 1 +...

Given,

$S = \left\{4, 6, 9\right\} \text{ and } T = \left\{9, 10, 11 \ldots . . 1000\right\}$

Also given $A\left\{{a}_{1} + {a}_{2} + \ldots . . + {a}_{k}: K \in N\right\} \text{ and } {a}_{i} \in S$

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 $x = y = 0 \text{ and } z = 1$ we get $a = 9$

Put $x = y = 1 \text{ and } z = 0$ we get $a = 10$

Put $x = 0, y = 2 \text{ and } z = 0$ we get $a = 12$

Put $x = 1, y = 0 \text{ and } z = 1$ we get $a = 13$

Put $x = 2, y = 1 \text{ and } z = o$ we get $a = 14$

Put $x = 0, y = 1 \text{ and } z = 1$ we get $a = 15$

. . .

Put $x = 1, y = 1 \text{ and } z = 1$ we get $a = 19$

Now from $20$ onwards we get all number as all number will be of type $4k, 4k + 1, 4k + 2 \text{ and } 4k + 3$ and from equation $1, 2 \text{ and } 3$ 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 $x, y, z \in W$

$∴ T - A = \left\{11\right\}$