In an arithmetic sequence, the explicit formula for the $n$-th term is given by:

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

where:

• $a_n$ is the $n$-th term,
• $a_1$ is the first term,
• $d$ is the common difference.

Given:

• The third term $a_3 = 1$,
• The ninth term $a_9 = 49$.

We can use these conditions to set up two equations.

1. For the third term: $a_3 = a_1 + 2d = 1$

2. For the ninth term: $a_9 = a_1 + 8d = 49$

Now, we solve this system of equations.

Step 1: Subtract the first equation from the second to eliminate $a_1$:

$(a_1 + 8d) - (a_1 + 2d) = 49 - 1$ $6d = 48$ $d = 8$

Step 2: Substitute $d = 8$ into the first equation:

$a_1 + 2(8) = 1$ $a_1 + 16 = 1$ $a_1 = -15$

Step 3: Write the explicit formula:

The explicit formula for the $n$-th term is:

$a_n = -15 + (n-1) \cdot 8$ $a_n = -15 + 8n - 8$ $a_n = 8n - 23$

Thus, the explicit formula for the arithmetic sequence is:

$a_n = 8n - 23$

Would you like a more detailed breakdown of any of the steps? Here are 5 related questions to consider:

1. How can you find the common difference in any arithmetic sequence?
2. What happens if the first term of an arithmetic sequence is unknown?
3. How do you find the sum of the first $n$ terms in an arithmetic sequence?
4. Can you solve a similar problem where the sixth term and twelfth term are given?
5. What is the connection between arithmetic sequences and linear functions?

Tip: In arithmetic sequences, the difference between consecutive terms remains constant, making them useful for identifying linear relationships.