, 2023

If a Gareth sequence begins 10, 8, then the 3rd number in the
sequence is $10-8=2$, the 4th is
$8-2=6$, the 5th is $6-2=4$, the 6th is $6-4=2$, the 7th is $4-2=2$, the 8th is $2-2=0$, the 9th is $2-0=2$, the 10th is $2-0=2$, and the 11th is $2-2=0$.

Thus, the resulting sequence is $10,8,2,6,4,2,2,0,2,2,0,\dots$.

The first 5 numbers in the sequence are $10,8,2,6,4$, the next 3 numbers are $2,2,0$, and this block of 3 numbers
appears to repeat.

Since each new number added to the end of this sequence is determined by
the two previous numbers in the sequence, then this block of 3 numbers
will indeed continue to repeat. (That is, since the block repeats once,
then it will continue repeating.)

The first 30 numbers of the sequence begins with the first 5 numbers,
followed by 8 blocks of $2,2,0$,
followed by one additional 2 (since $5+8\times 3+1=30$).

The sum of the first 5 numbers is $10+8+2+6+4=30$.

The sum of each repeating block is $2+2+0=4$, and so the sum of 8 such blocks
is $8\times 4=32$.

Thus, the sum of the first 30 numbers in the sequence is $30+32+2=64$.