# Problems from General term of an arithmetical progression

Find the general term of the arithmetical progression:

$$(1-\sqrt{2},1+\sqrt{2}, 1+3\sqrt{2}, 1+5\sqrt{2}, 1+7\sqrt{2}, \ldots)$$

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### 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: $$a_n=2\sqrt{2}n+1-3\sqrt{2}$$ Hide solution and development Find the fourth, eighth, hundredth and thirteenth terms in the arithmetical progression: $$\Big(\dfrac{1}{2}, \dfrac{3}{4}, 1, \dfrac{5}{4}, \dfrac{3}{2}, \ldots \Big)$$ See development and solution ### Development: 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 $$a_{113}=\dfrac{114}{4}=\dfrac{57}{2}=28+\dfrac{1}{2}$$$

### Solution:

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

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