make me an  summary explanation of lesson demonstrating the steps on how to find the 5th term of the arithmetic sequence ¹/6 ⅓ ½

### Summary Explanation: Finding the 5th Term of an Arithmetic Sequence

To find the 5th term of an arithmetic sequence, we can follow these steps:

#### Step 1: Identify the first term (\(a_1\))
The first term of the sequence is given as \( \frac{1}{6} \).

#### Step 2: Determine the common difference (\(d\))
The common difference \(d\) is found by subtracting the first term from the second term. 
For this sequence:
\[
d = \frac{1}{3} - \frac{1}{6} = \frac{2}{6} - \frac{1}{6} = \frac{1}{6}
\]

Verify by subtracting the second term from the third term:
\[
\frac{1}{2} - \frac{1}{3} = \frac{3}{6} - \frac{2}{6} = \frac{1}{6}
\]
So, \(d = \frac{1}{6}\).

#### Step 3: Use the formula for the nth term
The formula for the \(n\)th term of an arithmetic sequence is:
\[
a_n = a_1 + (n - 1) \times d
\]
Here, we want to find the 5th term (\(a_5\)).

#### Step 4: Substitute the values
Substitute \(a_1 = \frac{1}{6}\), \(d = \frac{1}{6}\), and \(n = 5\) into the formula:
\[
a_5 = \frac{1}{6} + (5 - 1) \times \frac{1}{6}
\]
\[
a_5 = \frac{1}{6} + 4 \times \frac{1}{6}
\]
\[
a_5 = \frac{1}{6} + \frac{4}{6} = \frac{1 + 4}{6} = \frac{5}{6}
\]

#### Step 5: Conclude the answer
The 5th term of the sequence is \( \frac{5}{6} \).

### Would you like details or have any questions?
#### Relative Questions:
1. How do you find the common difference in an arithmetic sequence?
2. What is the formula for the nth term of an arithmetic sequence?
3. How would the process change if the common difference was negative?
4. What happens to the sequence if the common difference is zero?
5. How do you determine if a sequence is arithmetic?

**Tip:** Always check the common difference between consecutive terms to ensure the sequence is arithmetic.

Discover how to find the 5th term of an arithmetic sequence with step-by-step instructions. This problem covers arithmetic sequence concepts and the nth term formula. Suitable for Grades 9-12.

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