smc

Let our first term be $a$ and our common difference be $d$. Thus, the first few terms of the sequence are $a$, $a+d$, $a+2d$, ... The sum of the first 5 terms is $$a+(a+d)+(a+2d)+...+(a+4d)=5a+10d$$ The sum of the first 10 terms is $$a+(a+d)+(a+2d)+...+(a+9d)=10a+45d$$ We are told that the latter is four times the former. Hence, $$4(5a+10d)=10a+45d$$ $$20a+40d=10a+45d$$ $$10a=5d$$ $$\frac{a}{d}=\frac{1}{2}$$ Therefore, our answer is $\boxed{1:2}$