Explanation:
Okay, so I need to find the integer value of x in the arithmetic sequence 3², x, 3⁴. Let me start by recalling what an arithmetic sequence is. An arithmetic sequence is a sequence of numbers where the difference between consecutive terms is constant. That means if I have three terms in an arithmetic sequence, the middle term is the average of the first and third terms.

First, let me write down the given terms. The first term is 3 squared, which is 9. The second term is x, which is what we need to find. The third term is 3 to the fourth power, which is 81. So the sequence is 9, x, 81.

Since it's an arithmetic sequence, the difference between the first and second term should be the same as the difference between the second and third term. Let me denote the common difference as d. So, the difference between the first and second term is x - 9, and the difference between the second and third term is 81 - x. Since these differences are equal, I can set up the equation:

x - 9 = 81 - x

Now, I need to solve for x. Let me do that step by step.

First, I can add x to both sides of the equation to get rid of the -x on the right side.

x - 9 + x = 81 - x + x

Simplifying both sides:

2x - 9 = 81

Now, I need to isolate x. Let me add 9 to both sides to get rid of the -9 on the left side.

2x - 9 + 9 = 81 + 9

Simplifying:

2x = 90

Now, divide both sides by 2 to solve for x:

x = 90 / 2

x = 45

Wait, so x is 45? Let me check if that makes sense. Let me verify if 9, 45, 81 is indeed an arithmetic sequence.

The difference between the first and second term is 45 - 9 = 36. The difference between the second and third term is 81 - 45 = 36. Yes, both differences are 36, so that checks out.

But let me think again. The problem says "the integer value of x". Well, 45 is an integer, so that's good. Is there another way to approach this problem? Maybe using the formula for the nth term of an arithmetic sequence?

In an arithmetic sequence, the nth term is given by a_n = a_1 + (n - 1)d, where a_1 is the first term and d is the common difference.

In this case, we have three terms. Let me label them:

First term (n=1): a_1 = 9

Second term (n=2): a_2 = x

Third term (n=3): a_3 = 81

So, using the formula for the second term:

a_2 = a_1 + (2 - 1)d = a_1 + d

Which gives x = 9 + d

Similarly, for the third term:

a_3 = a_1 + (3 - 1)d = a_1 + 2d

Which gives 81 = 9 + 2d

Solving for d:

81 - 9 = 2d

72 = 2d

d = 36

Then, substituting back into the equation for x:

x = 9 + d = 9 + 36 = 45

Same result. So that confirms it again.

Alternatively, since in an arithmetic sequence the middle term is the average of the first and third term. So, x = (first term + third term)/2.

So, x = (9 + 81)/2 = 90/2 = 45. Yep, same answer.

Therefore, regardless of the method I use, I get x = 45.

But let me just make sure that there isn't a trick or something I'm missing here. The problem says "arithmetic sequence", so it's not a geometric sequence or anything else. The terms are 3², x, 3⁴. So 3 squared is 9, 3 to the fourth is 81. So the sequence is 9, x, 81.

Is there a possibility that the question is referring to something else? Like maybe the exponents form an arithmetic sequence? Wait, but the problem says "the arithmetic sequence 3², x, 3⁴". So the terms themselves are 3², x, 3⁴. So the terms are numbers, not exponents. So the sequence is numerical.

So, for example, if it was a geometric sequence, then x would be 3^3 = 27, but since it's arithmetic, it's different.

But just to check, if someone thought that maybe the exponents are in arithmetic sequence, but that would be a different problem. Let me see. If the exponents were in arithmetic sequence, then 2, x, 4. Then the common difference would be (x - 2) and (4 - x). So x - 2 = 4 - x => 2x = 6 => x = 3. Then the exponents would be 2, 3, 4, but the terms would be 3², 3³, 3⁴, which is a geometric sequence. But the problem says arithmetic sequence, so that's not the case.

Therefore, the original problem is about the terms being in arithmetic sequence, not the exponents. So the answer is 45.

Another way to think about it: the difference between the first and third term is 81 - 9 = 72. In an arithmetic sequence, the difference between the first and third term is 2d. Therefore, 2d = 72 => d = 36. Then, the second term is first term + d = 9 + 36 = 45.

All methods lead to the same answer. Therefore, I can be confident that x is 45.

Just to recap:

Given arithmetic sequence: 9, x, 81

Common difference d = x - 9 = 81 - x

Solving gives x = 45.

Yes, that's straightforward.

I think that's all. I don't see any mistakes in my reasoning. So the integer value of x is 45.

**Final Answer**
The integer value of $ x $ is \boxed{45}.

To determine the integer value of $ x $ in the arithmetic sequence $ 3^2, x, 3^4 $, we begin by identifying the numerical values of the given terms:

- $ 3^2 = 9 $
- $ 3^4 = 81 $

So the sequence becomes:
$$
9, \ x, \ 81
$$

---

### Step 1: Use the property of arithmetic sequences

In an arithmetic sequence, the difference between consecutive terms is constant. That is, the difference between the first and second term is the same as the difference between the second and third term:

$$
x - 9 = 81 - x
$$

---

### Step 2: Solve the equation

Add $ x $ to both sides:

$$
x - 9 + x = 81
$$

$$
2x - 9 = 81
$$

Add 9 to both sides:

$$
2x = 90
$$

Divide both sides by 2:

$$
x = 45
$$

---

### Step 3: Verify the result

Check that the sequence $ 9, 45, 81 $ is indeed arithmetic:

- $ 45 - 9 = 36 $
- $ 81 - 45 = 36 $

Since the common difference is consistent, the sequence is arithmetic, and the value of $ x $ is correct.

---

### Final Answer

$$
\boxed{45}
$$

[Assistant]
Explanation:
Okay, so I need to find the integer value of x in the arithmetic sequence 3², x, 3⁴. Let me start by recalling what an arithmetic sequence is. An arithmetic sequence is a sequence of numbers where the difference between consecutive terms is constant. That means if I have three terms in an arithmetic sequence, the middle term is the average of the first and third terms.

First, let me write down the given terms. The first term is 3 squared, which is 9. The second term is x, which is what we need to find. The third term is 3 to the fourth power, which is 81. So the sequence is 9, x, 81.

Since it's an arithmetic sequence, the difference between the first and second term should be the same as the difference between the second and third term. Let me denote the common difference as d. So, the difference between the first and second term is x - 9, and the difference between the second and third term is 81 - x. Since these differences are equal, I can set up the equation:

x - 9 = 81 - x

Now, I need to solve for x. Let me do that step by step.

First, I can add x to both sides of the equation to get rid of the -x on the right side.

x - 9 + x = 81 - x + x

Simplifying both sides:

2x - 9 = 81

Now, I need to isolate x. Let me add 9 to both sides to get rid of the -9 on the left side.

2x - 9 + 9 = 81 + 9

Simplifying:

2x = 90

Now, divide both sides by 2 to solve for x:

x = 90 / 2

x = 45

Wait, so x is 45? Let me check if that makes sense. Let me verify if 9, 45, 81 is indeed an arithmetic sequence.

The difference between the first and second term is 45 - 9 = 36. The difference between the second and third term is 81 - 45 = 36. Yes, both differences are 36, so that checks out.

But let me think again. The problem says "the integer value of x". Well, 45 is an integer, so that's good. Is there another way to approach this problem? Maybe using the formula for the nth term of an arithmetic sequence?

In an arithmetic sequence, the nth term is given by a_n = a_1 + (n - 1)d, where a_1 is the first term and d is the common difference.

In this case, we have three terms. Let me label them:

First term (n=1): a_1 = 9

Second term (n=2): a_2 = x

Third term (n=3): a_3 = 81

So, using the formula for the second term:

a_2 = a_1 + (2 - 1)d = a_1 + d

Which gives x = 9 + d

Similarly, for the third term:

a_3 = a_1 + (3 - 1)d = a_1 + 2d

Which gives 81 = 9 + 2d

Solving for d:

81 - 9 = 2d

72 = 2d

d = 36

Then, substituting back into the equation for x:

x = 9 + d = 9 + 36 = 45

Same result. So that confirms it again.

Alternatively, since in an arithmetic sequence the middle term is the average of the first and third term. So, x = (first term + third term)/2.

So, x = (9 + 81)/2 = 90/2 = 45. Yep, same answer.

Therefore, regardless of the method I use, I get x = 45.

But let me just make sure that there isn't a trick or something I'm missing here. The problem says "arithmetic sequence", so it's not a geometric sequence or anything else. The terms are 3², x, 3⁴. So 3 squared is 9, 3 to the fourth is 81. So the sequence is 9, x, 81.

Is there a possibility that the question is referring to something else? Like maybe the exponents form an arithmetic sequence? Wait, but the problem says "the arithmetic sequence 3², x, 3⁴". So the terms themselves are 3², x, 3⁴. So the terms are numbers, not exponents. So the sequence is numerical.

So, for example, if it was a geometric sequence, then x would be 3^3 = 27, but since it's arithmetic, it's different.

But just to check, if someone thought that maybe the exponents are in arithmetic sequence, but that would be a different problem. Let me see. If the exponents were in arithmetic sequence, then 2, x, 4. Then the common difference would be (x - 2) and (4 - x). So x - 2 = 4 - x => 2x = 6 => x = 3. Then the exponents would be 2, 3, 4, but the terms would be 3², 3³, 3⁴, which is a geometric sequence. But the problem says arithmetic sequence, so that's not the case.

Therefore, the original problem is about the terms being in arithmetic sequence, not the exponents. So the answer is 45.

Another way to think about it: the difference between the first and third term is 81 - 9 = 72. In an arithmetic sequence, the difference between the first and third term is 2d. Therefore, 2d = 72 => d = 36. Then, the second term is first term + d = 9 + 36 = 45.

All methods lead to the same answer. Therefore, I can be confident that x is 45.

Just to recap:

Given arithmetic sequence: 9, x, 81

Common difference d = x - 9 = 81 - x

Solving gives x = 45.

Yes, that's straightforward.

I think that's all. I don't see any mistakes in my reasoning. So the integer value of x is 45.

**Final Answer**
The integer value of $ x $ is \boxed{45}.

To determine the integer value of $ x $ in the arithmetic sequence $ 3^2, x, 3^4 $, we begin by identifying the numerical values of the given terms:

- $ 3^2 = 9 $
- $ 3^4 = 81 $

So the sequence becomes:
$$
9, \ x, \ 81
$$

---

### Step 1: Use the property of arithmetic sequences

In an arithmetic sequence, the difference between consecutive terms is constant. That is, the difference between the first and second term is the same as the difference between the second and third term:

$$
x - 9 = 81 - x
$$

---

### Step 2: Solve the equation

Add $ x $ to both sides:

$$
x - 9 + x = 81
$$

$$
2x - 9 = 81
$$

Add 9 to both sides:

$$
2x = 90
$$

Divide both sides by 2:

$$
x = 45
$$

---

### Step 3: Verify the result

Check that the sequence $ 9, 45, 81 $ is indeed arithmetic:

- $ 45 - 9 = 36 $
- $ 81 - 45 = 36 $

Since the common difference is consistent, the sequence is arithmetic, and the value of $ x $ is correct.

---

### Final Answer

$$
\boxed{45}
$$

Solution: The term $x$ is simply the average of $3^2 = 9$ and $3^4 = 81$, which is $(9 + 81)/2 = 90/2 = \boxed{45}$.