From the image, it looks like we are dealing with arithmetic sequence problems where we need to solve for a specific term, the common difference, or the number of terms in a sequence. Here's a breakdown for each problem shown:

Problem 28: Given the first term $a_1 = 5$ and the 10th term $a_{10} = 85$, we need to find the common difference $d$.

Using the formula for the nth term of an arithmetic sequence, $a_n = a_1 + (n-1) \cdot d$, we can plug in the known values: $a_{10} = 5 + (10-1) \cdot d = 85$

Problem 29: Given the first term $a_1 = 4$ and the 12th term $a_{12} = 26$, we need to find the common difference $d$ as well.

Using the same formula: $a_{12} = 4 + (12-1) \cdot d = 26$

Problem 30: Given the first term $a_1 = -1$, the common difference $d = 5$, and the nth term $a_n = 34$, we need to find the number of terms $n$.

Again, using the nth term formula: $a_n = -1 + (n-1) \cdot 5 = 34$

Let's solve these equations to find the values of $d$ and $n$.Here are the solutions for each problem:

Problem 28: The common difference $d$ is $\frac{80}{9}$ or approximately $8.89$.

Problem 29: The common difference $d$ is $2$.

Problem 30: The sequence has $8$ terms.

Would you like to understand more about any of these problems or do you have any other questions?

Here are some follow-up topics you might consider:

1. How does changing the common difference $d$ affect the sequence?
2. What would the sequence look like if you extended it to more terms?
3. How do you find the sum of the first $n$ terms in an arithmetic sequence?
4. Can you have a negative common difference? What does it signify?
5. How does the formula for the nth term derive from the sequence definition?

Tip: Always verify your arithmetic sequence results by substituting back into the original formula to ensure consistency and correctness.