C2 Arithmatic sequences

board?

This question part ii) has gotten me quite confused. Here's my approach:
So I basically made u1,u2 a sequence knowing that it repeats every 3 values.
So we can start at u1 first.
a = 1 d = 3
1+(n-1)3=200
n=199/3 we can see that a decimal is received so it's not possible
Then attempt with u2 starting so a=2 d = 3
2+(n-1)3=200
here 198/3=66 so it's divisible and therefore the answer is -1.
This actually gets me the correct answer but I have no clue why. I just did this of half guessing. Could anyone please explain why this arrives me to the correct answer? Thanks

Because your sequence given in question is $\{ 2,-1,\frac{1}{2},2,-1,\frac{1}{2},...\}$

This means that $\displaystyle -1=u_2=u_5=u_8=...=u_{2+3k}=...$ for $k \in \mathbb{N}$

Setting $2+3k=200$ gives you a natural number for $k$, which fits the above, so $u_{200}=-1$

Because your sequence given in question is $\{ 2,-1,\frac{1}{2},2,-1,\frac{1}{2},...\}$

This means that $\displaystyle -1=u_2=u_5=u_8=...=u_{2+3k}=...$ for $k \in \mathbb{N}_0$

Setting $2+3k=200$ gives you a natural number for $k$, which fits the above, so $u_{200}=-1$

So instead of using an arithmetic sequence I should just use nth term?
And thanks that clarifies it

What would you use as d in the nth term formula?

If I were to do the question, I would recognise that since the sequence repeats after every 3 terms, then I'd look for the closest number to 200 that is a multiple of 3 - so that would be 201. Meaning $u_{201}=\frac{1}{2}$ so go back one and you get $u_{200}=-1$

Why must it be a half?

Because if it repeats after every 3rd term, this means that every 3rd term is the last number in the cycle. Which is 1/2.

To expand on the above just by a bit and shed a bit more light,

From your sequence $\{ 2,-1,\frac{1}{2},2,-1,\frac{1}{2},...\}$

You get:
$2=u_1=u_4=u_7=...=u_{1+3(k-1)}=...$

$-1=u_2=u_5=u_8=...=u_{2+3(k-1)}=...$

$\frac{1}{2}=u_3=u_6=u_9=...=u_{3+3(k-1)}=...$

...for $k \in \mathbb{N}$


Now if $u_{200}=2$ THEN there exists a natural number $K$ such that $1+3(K-1)=200$

If $u_{200}=-1$ THEN there exists a natural number $K$ such that $2+3(K-1)=200$

If $u_{200}=\frac{1}{2}$ THEN there exists a natural number $K$ such that $3+3(K-1)=200$


So your method is sort of a trial and error as you're looking through these possibilities to see which one is satisfied, it is not recommended if the cycle is larger.



If I were to do the question, I would recognise that since the sequence repeats after every 3 terms (every 3rd term is 1/2), then I'd look for the closest number to 200 that is a multiple of 3 - so that would be 201. Meaning $u_{201}=\frac{1}{2}$ so go back one and you get $u_{200}=-1$