From the image, the problem provides two terms of a sequence:

• $a_3 = 8$
• $a_6 = \frac{29}{4}$

This likely refers to an arithmetic or geometric sequence, where we need to find further terms, the common difference (in arithmetic sequences), or the common ratio (in geometric sequences), or determine the formula for the sequence.

Let's first assume it's an arithmetic sequence, which follows the form: $a_n = a_1 + (n-1) \cdot d$ where $a_n$ is the $n$-th term, $a_1$ is the first term, $n$ is the term number, and $d$ is the common difference.

We have two equations from the given terms:

1. $a_3 = a_1 + 2d = 8$
2. $a_6 = a_1 + 5d = \frac{29}{4}$

We can solve these equations to find $a_1$ and $d$. Let's proceed with that.

Step 1: Express $a_1$ in terms of $d$ from the first equation: $a_1 = 8 - 2d$

Step 2: Substitute into the second equation: $8 - 2d + 5d = \frac{29}{4}$ $8 + 3d = \frac{29}{4}$

Step 3: Solve for $d$: First, subtract 8 from both sides: $3d = \frac{29}{4} - 8 = \frac{29}{4} - \frac{32}{4} = -\frac{3}{4}$ Now, divide both sides by 3: $d = -\frac{3}{4} \times \frac{1}{3} = -\frac{1}{4}$

Step 4: Find $a_1$: Substitute $d = -\frac{1}{4}$ into $a_1 = 8 - 2d$: $a_1 = 8 - 2\left(-\frac{1}{4}\right) = 8 + \frac{1}{2} = \frac{16}{2} + \frac{1}{2} = \frac{17}{2}$

So, the sequence is an arithmetic sequence with:

• First term $a_1 = \frac{17}{2}$
• Common difference $d = -\frac{1}{4}$

If you need more details or clarifications, let me know!

Related Questions:

1. What is the formula for the $n$-th term of this arithmetic sequence?
2. How do you find the sum of the first $n$ terms in an arithmetic sequence?
3. What is the 10th term of this sequence?
4. Can you show how to solve this if it were a geometric sequence instead of arithmetic?
5. How does the common difference affect the behavior of the sequence?

Tip: Always double-check if a sequence is arithmetic or geometric by testing the ratio or difference between consecutive terms!