General term of an arithmetical progression

Let's see what the difference is $$d=(1+\sqrt{2})-(1-\sqrt{2})=\sqrt{2}+\sqrt{2}=2\sqrt{2}$$ And, as the first term is $a_1=1-\sqrt{2}$, we know that:

$$a_n=(1-\sqrt{2})+(n-1)\cdot 2 \cdot \sqrt{2}$$

Arranging this expression we have:

$$a_n=(1-\sqrt{2})+2\sqrt{2}(n-1)=1-\sqrt{2}-2\sqrt{2}+2\sqrt{2}\cdot n=2\sqrt{2}n+1-3\sqrt{2}$$

The difference is $d=\dfrac{3}{4}-\dfrac{1}{2}=\dfrac{1}{4}$, and as $a_1=\dfrac{1}{2}$, we have that $$a_n=\dfrac{1}{2}+\dfrac{1}{4}(n-1)=\dfrac{n}{4}+\dfrac{1}{4}=\dfrac{n+1}{4}$$

So:

$$a_4=\dfrac{4+1}{4}=\dfrac{5}{4}, \ a_8=\dfrac{9}{4}$$ and $$a_{113}=\dfrac{114}{4}=\dfrac{57}{2}=28+\dfrac{1}{2}$$