<Question>
If the fourth term of an arithmetic sequence is $200$ and the eighth term is $500$, what is the sixth term?
</Question>

<think>
<Plan_1>
Okay, let's see. I need to find the sixth term of an arithmetic sequence where the fourth term is 200 and the eighth term is 500. Hmm, arithmetic sequences... right, they have a common difference between consecutive terms. Let me recall the formula for the nth term of an arithmetic sequence. I think it's a_n = a_1 + (n-1)d, where a_1 is the first term and d is the common difference.

So, if the fourth term is 200, that would be a_4 = a_1 + 3d = 200. And the eighth term is a_8 = a_1 + 7d = 500. So I have two equations here:

1. a_1 + 3d = 200
2. a_1 + 7d = 500

I need to solve these two equations to find a_1 and d. Then, once I have those, I can find the sixth term, which is a_6 = a_1 + 5d.

Let me subtract the first equation from the second to eliminate a_1.

<prompt_1.1> Subtract the first equation from the second to find the common difference \( d \). </prompt_1.1>
-----

Alternatively, maybe there's another way to approach this problem without solving for a_1. Let me think. Since the terms are in an arithmetic sequence, the difference between terms equidistant from a given term can be related. For example, the eighth term and the fourth term are four terms apart. The sixth term is two terms after the fourth term.

<prompt_1.2> Explore an alternative method to find the sixth term without solving for \( a_1 \). </prompt_1.2>
</Plan_1>

<Execution_1>
<execution_1.1>
So, (a_1 + 7d) - (a_1 + 3d) = 500 - 200. Simplifying the left side: a_1 - a_1 + 7d - 3d = 4d. And the right side is 300. So 4d = 300. Therefore, d = 300 / 4 = 75.
</execution_1.1>

<execution_1.2>
Wait, maybe there's a way to find the sixth term directly from the fourth and eighth terms by using the concept of arithmetic sequences. Let's see. The fourth term is 200, the eighth term is 500. The difference between these two terms is 500 - 200 = 300. Since there are 8 - 4 = 4 terms apart, the common difference per term would be 300 / 4 = 75. Wait, that's the same as before. Then, to get from the fourth term to the sixth term, which is two terms away, we add 2*75 = 150, so 200 + 150 = 350. That's the same answer.
</execution_1.2>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Now that I have d, I can substitute back into one of the original equations to find a_1. Let's use the first equation: a_1 + 3*75 = 200.
So 3*75 is 225. Therefore, a_1 = 200 - 225 = -25.

Hmm, a negative first term? That's okay, arithmetic sequences can have negative terms. Let me check if this makes sense with the eighth term.
Using a_1 = -25 and d = 75, the eighth term would be a_1 + 7d = -25 + 7*75. 7*75 is 525, so -25 + 525 = 500.
-----

Based on execution_1.2:
Alternatively, maybe I can use the concept of average? Since in an arithmetic sequence, the term halfway between two terms is the average of those two terms. But the sixth term is halfway between the fourth and eighth terms? Let's see.
The fourth term is 200, eighth is 500. The sixth term is two terms after the fourth, and two terms before the eighth. So, the sixth term is the average of the fourth and eighth terms? Wait, let me check.

<prompt_2.1> Verify if the sixth term can be found using the average of the fourth and eighth terms. </prompt_2.1>
</Plan_2>

<Execution_2>
<execution_2.1>
If the sixth term is the average of the fourth and eighth terms, then (200 + 500)/2 = 700/2 = 350. That's the same answer as before. So that works too. So maybe there's a shortcut here: since the sixth term is equidistant from the fourth and eighth terms, it's the average of those two terms. That would save a step. But either way, whether by solving for a_1 and d, or by recognizing that the sixth term is the average of the fourth and eighth terms, we get the same result.
</execution_2.1>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Wait, let me confirm why taking the average works here. In an arithmetic sequence, the terms are linearly increasing (or decreasing) with a constant difference. Therefore, any term that is symmetrically placed between two terms can be found by averaging. For example, if you have two terms a_m and a_n where m < k < n and k is equidistant from m and n, then a_k is the average of a_m and a_n.

<prompt_3.1> Verify the method of finding the sixth term by averaging the fourth and eighth terms. </prompt_3.1>

So, the sixth term is 350. Let me just verify once more with the original method.

<prompt_3.2> Verify the sixth term using the original method of solving for \( a_1 \) and \( d \). </prompt_3.2>
</Plan_3>

<Execution_3>
<execution_3.1>
That makes sense because the difference between a_k and a_m is (k - m)d, and between a_n and a_k is (n - k)d. If k is the midpoint, then (k - m) = (n - k), so k = (m + n)/2. Therefore, a_k = a_m + (k - m)d. But since d is constant, maybe there's a relation here. Let's test it with the given numbers.

The fourth term is 200, the eighth term is 500. The sixth term is the third term after the fourth, and the second term before the eighth. Wait, no. The sixth term is two terms after the fourth (4th, 5th, 6th) and three terms before the eighth (6th, 7th, 8th). Wait, that's not symmetric. Wait, maybe my initial thought was wrong. Let's see.

Wait, the sixth term is at position 6. The fourth term is 4, eighth is 8. The distance from 4 to 6 is 2, and from 6 to 8 is 2. So the sixth term is exactly in the middle between the fourth and eighth terms. Therefore, it's the average. So even though the number of intervals between 4 and 6 is 2 intervals (each of d), and between 6 and 8 is another 2 intervals. Therefore, the term at 6 is the average of 4 and 8. Therefore, (200 + 500)/2 = 350. So that's a quicker way.
</execution_3.1>

<execution_3.2>
We found a_1 = -25, d = 75. Then the sixth term is a_1 + 5d = -25 + 5*75. 5*75 is 375. 375 - 25 is 350.
</execution_3.2>
</Execution_3>

<Final_answer>
Therefore, the sixth term of the arithmetic sequence is $\boxed{350}$.
</Final_answer>
</think>