Let S - 4- 6- 9 and T - 9- 10- 11- -1000- IfA-a1-a2-ak-k- N-a1-a2-a3-ak- S- then the sum of all the elements in the set T - A is equal to -

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nth Term of A.P

Let S = 4, 6,...

Question

Let $S$ = { $4, 6, 9$ } and $T$ = { $9, 10, 11, . . ., 1000$ }. If
$A$ ={ $a_{1} + a_{2} + \dots + a_{k} : k \in N, a_{1}, a_{2}, a_{3}, \dots, a_{k} \in S$ }, then the sum of all the elements in the set $T - A$ is equal to .

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Solution

Here $S = {4, 6, 9}$

And $T = {9, 10, 11, \dots, 1000}$ .

We have to find all numbers in the form of
$4 x + 6 y + 9 z$ , where $x, y, z \in {0, 1, 2, \dots}$ .

If $a$ and $b$ are coprime number then the least
number from which all the number more than or equal to it can be express as $a x + b y$ where $x, y \in {0, 1, 2, \dots}$ is $(a - 1) \cdot (b - 1)$ .

Then for $6 y + 9 z = 3 (2 y + 3 z)$

All the number from $(2 - 1) \cdot (3 - 1) = 2$ and above can be express as $2 x + 3 z ($ say $t$ ).

Now $4 x + 6 y + 9 z = 4 x + 3 (t + 2)$

$= 4 x + 3 t + 6$

again by same rule $4 x + 3 t$ , all the number from $(4 - 1) (3 - 1) = 6$ and above can be express from $4 x + 3 t$ .

Then $4 x + 6 y + 9 z$ express all the numbers from $12$ and above.

again $9$ and $10$ can be express in form $4 x + 6 y + 9 z$ .

Then set $A = {9, 10, 12, 13, \dots, 1000}$ .

Then $T - A = {11}$

Only one element $11$ is there.

Sum of elements of $T - A = 11$

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a_{1}, a_{2}, a_{3}, . . . ., a_{11}

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\frac{a_{1} + a_{2} + . . . + a_{11}}{11}

is equal to

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Let S = {4,6,9} and T = {9,10,11,...,1000}. If A={a1+a2+⋯+ak:k∈ N,a1,a2,a3,⋯,ak∈S}, then the sum of all the elements in the set T−A is equal to .

Let $S$ = { $4, 6, 9$ } and $T$ = { $9, 10, 11, . . ., 1000$ }. If
$A$ ={ $a_{1} + a_{2} + \dots + a_{k} : k \in N, a_{1}, a_{2}, a_{3}, \dots, a_{k} \in S$ }, then the sum of all the elements in the set $T - A$ is equal to .