Problems from General term of an arithmetical progression

Development:

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:

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

Solution:

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

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Development:

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

So:

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

Solution:

$a_4={\displaystyle \frac{5}{4}},a_8={\displaystyle \frac{9}{4}}$ and $a_{113}={\displaystyle \frac{57}{2}}$

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