Residue of a sequence

As we have to find the elements that are a multiple of 8, we work with modulo 8.

$a_1 \equiv 1 \pmod 8$

$a_2 \equiv 2 \pmod 8$

$a_3 \equiv 3\times 2+5\times 1 \equiv 11 \equiv 3 \pmod 8$

$a_4 \equiv 3\times 3+5\times 2 \equiv 19 \equiv 3 \pmod 8$

$a_5 \equiv 3\times 3+5\times 3 \equiv 24 \equiv 0 \pmod 8$

$a_6 \equiv 3\times 0+5\times 2 \equiv 7 \pmod 8$

$a_7 \equiv 3\times 7+5\times 0 \equiv 5 \pmod 8$

$a_8 \equiv 3\times 5+5\times 7 \equiv 2 \pmod 8$

$a_9 \equiv 3\times 2+5\times 5 \equiv 7 \pmod 8$

$a_{10} \equiv 3\times 7+5\times 2 \equiv 7 \pmod 8$

$a_{11} \equiv 3\times 7+5\times 7 \equiv 0 \pmod 8$

$a_{12} \equiv 3\times 0+5\times 7 \equiv 3 \pmod 8$

$a_{13} \equiv 3\times 3+5\times 0 \equiv 1 \pmod 8$

$a_{14} \equiv 3\times 1+5\times 3 \equiv 2 \pmod 8$

We notice that the residues of $a_n\pmod{8}$ repeat with periodicity 12.

There are two multiples of 8 in a period of 12. hence there are 200 multiples of 8 in $\{ a_1,a_2,a_3......a_{1200} \}$ .

Hence, the probability is $\dfrac{200}{1200}=\dfrac{1}{6}$

Therefore, $a=1,\,\,b=6 \Rightarrow a+b=\fbox{7}$ .

By examining the sequence modulo 8, we see it is periodic with period 12: $1,2,3,3,0,7,5,2,7,7,0,3,1,2,\dots$ Out of 12 elements in any period, 2 are 0 modulo 8. Since the set consists of $12\cdot 100$ consecutive elements, exactly $2/12 = 1/6$ of them will be 0 modulo 8, and so the answer is $1+6=\boxed{7}$ .