The 3rd term of an arithmetic sequence is 14 and the 7th term is 26. Find the 12th term.

To solve this problem, we can use the formula for the \( n \)-th term of an arithmetic sequence:

\[
a_n = a_1 + (n-1) \cdot d
\]

where:
- \( a_n \) is the \( n \)-th term,
- \( a_1 \) is the first term,
- \( d \) is the common difference,
- \( n \) is the term number.

Given:
- The 3rd term, \( a_3 = 14 \),
- The 7th term, \( a_7 = 26 \).

### Step 1: Set up equations for the given terms
Using the general formula for the 3rd term:
\[
a_3 = a_1 + 2d = 14
\]
Using the general formula for the 7th term:
\[
a_7 = a_1 + 6d = 26
\]

### Step 2: Subtract the first equation from the second
\[
(a_1 + 6d) - (a_1 + 2d) = 26 - 14
\]
\[
4d = 12
\]
\[
d = 3
\]

### Step 3: Find the first term \( a_1 \)
Substitute \( d = 3 \) into the equation for the 3rd term:
\[
a_1 + 2(3) = 14
\]
\[
a_1 + 6 = 14
\]
\[
a_1 = 8
\]

### Step 4: Calculate the 12th term
Using the general formula for the 12th term:
\[
a_{12} = a_1 + 11d
\]
Substitute \( a_1 = 8 \) and \( d = 3 \):
\[
a_{12} = 8 + 11(3) = 8 + 33 = 41
\]

So, the 12th term is \( \boxed{41} \).

Do you want more details or have any questions? Here are some related questions to explore:

1. How do you find the common difference if the first and last terms are given?
2. What is the sum of the first 12 terms of this arithmetic sequence?
3. Can you derive the formula for the sum of the first \( n \) terms in an arithmetic sequence?
4. How would you find the \( n \)-th term if you were given two non-consecutive terms?
5. What happens to the sequence if the common difference \( d \) is negative?

**Tip:** In arithmetic sequences, knowing just two terms allows you to find the common difference and the first term, which in turn lets you determine any term in the sequence.

Learn how to find the 12th term of an arithmetic sequence given the 3rd term as 14 and the 7th term as 26. This problem involves using the formula for the nth term of an arithmetic sequence and solving for the common difference and first term. Suitable for students in Grades 9-10.

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