This is a cute problem, if there are no answers soon I'll take a stab at it.

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Just a note, the way the question is worded is not unusual for a maths competition question. But the solver just needs to be a bit careful in processing the algebra.

For example, the "tenth term" is *not* G(10) ;-)

G(0)=a
G(1)=b
G(2)=a+b
G(3)=a+2b
G(4) =2a+3b
G(5)=3a+5b
G(6)=5a+8b
...
G(n)=F(n-1)×a+F(n)×b

for F(n):
F(0)=0
F(1)=1
F(2)=1
F(3)=2
F(4)=3
F(5)=5
...
I won't give the formula for the usual Fibonacci sequence here since you could just Google it your self. I'm pretty sure I could extract it from what I've done so far but It's not necessary for what I'm trying to point out. 

So for 
G(10) it's 34×a+55×b
And since G(10) should be 2018 you get the equation
2018=34a+55b.

a+b minimal if b almost equal a
b=a+c
2018=89a+55c

So a pretty good solution would be:
2017+1=89a+55c
2017=89a
1=55c
so a would be 2017/188 and b 2017/188+1/55
An even better solution would be:
2017,9+0,1=89a+55c
2017,9=89a
0,1=55c
so a=20179/1880 and b=20179/1880+1/550
and so on
next would be a=201799/18800 b=201799/18800+1/5500
And so on.

Hope I didn't make any mistakes somewhere.

All right. If we start with G(0) as first therm it would be G(9) which would be after the formula I pointed out in another comment
2018=21a+34b 
again 2018 mod 55 is 38 2018-38=1980 1980/55=36
so the a has to be 36 or less.
2018-21a=34b
1009-21a/2=17b
a/2=c-> cmax=18
(1009-21c)/17=b

for a+b minimal a has to be as small as posible since b has a bigger impact (2ax=34 x has to be bigger than 1.5)
So you can either make a table or test for each number for c beginning at 1 and stopping as soon as b is an integer.
I dicided to go with the last option. I won't post the boring calculations.
It turns out it works for c=10 and so b is 47 and a is 20.
I checked even further and it turns out there is no other possible pair smaler than c=18 and if c larger 18 a smaller b would not apply.

The 10th term in the sequence is 21a+34b = 2018 or 21(a+b) + 13b = 2018.  Since 21 x 96 = 2016, which is 2018-2, our objective is to find b that will make 13b-2 divisible by 21.  The smallest positive integer b satisfying this condition is 5.  Therefore, 21(a+b) + 13b = 21(a+5) + 13x5 = 21a + 170 = 2018, i.e. 21a = 1848 or a = 88. 

So the smallest possible value of (a+b) should be 88 + 5 = 93

G(10)=2018=34*G(0)+55*G(1)
=>G(0)=(2018-55*G(1))/34;
=>G(0)=27;G(1)=20;
But 27>20 => there are no solution