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Let's solve each of the problems (26 to 30) one by one briefly:

### Problem 26:
**Determine the sum of three terms of a geometric progression, knowing that the sum of the first two terms is 8, and the difference between the second and the third term is 4.**

Let the terms of the geometric progression be \( a \), \( ar \), and \( ar^2 \). We are given:
1. \( a + ar = 8 \) → \( a(1 + r) = 8 \)
2. \( ar^2 - ar = 4 \) → \( ar(r - 1) = 4 \)

From the first equation, solve for \( a \):
\[ a = \frac{8}{1 + r} \]

Substitute this into the second equation:
\[
\frac{8r(r - 1)}{1 + r} = 4
\]

Solve for \( r \), then substitute back to find \( a \). This will give the terms and their sum.

---

### Problem 27:
**Calculate the fifth term of an arithmetic progression, knowing that the first term is 7 and the second term is 9.**

In an arithmetic progression, the \( n \)-th term is given by:
\[ a_n = a_1 + (n-1)d \]

Given:
- \( a_1 = 7 \)
- \( a_2 = 9 \)

We know:
\[ a_2 = a_1 + d \]
Thus, \( 9 = 7 + d \), so \( d = 2 \).

The fifth term:
\[
a_5 = 7 + (5-1) \times 2 = 7 + 8 = 15
\]

So, the fifth term is 15.

---

### Problem 28:
**Determine the ratio of the geometric progression \( (b_n)_{n \geq 1} \), knowing that \( b_1 = 3 \) and \( b_2 - b_1 = 3 \).**

We are given:
- \( b_1 = 3 \)
- \( b_2 - b_1 = 3 \)

Since it's a geometric progression:
\[ b_2 = b_1 \cdot r \]

Thus:
\[ 3r - 3 = 3 \]
\[
3r = 6 \quad \Rightarrow \quad r = 2
\]

So, the ratio of the geometric progression is \( r = 2 \).

---

### Problem 29:
**Prove that for the sequence with the general term \( a_n = 2n + 3 \), the relation \( a_{n+1} - a_n = 2 \) holds for all \( n \in \mathbb{N} \).**

The general term is \( a_n = 2n + 3 \).

Now, calculate:
\[ a_{n+1} = 2(n+1) + 3 = 2n + 2 + 3 = 2n + 5 \]

Now find the difference:
\[ a_{n+1} - a_n = (2n + 5) - (2n + 3) = 2 \]

Thus, the relation \( a_{n+1} - a_n = 2 \) is proven for all \( n \in \mathbb{N} \).

---

### Problem 30:
**Determine the real number \( x \), knowing that the numbers \( x-1 \), \( x+1 \), and \( 2x-1 \) are consecutive terms of an arithmetic progression.**

In an arithmetic progression, the difference between consecutive terms is constant. So, we can write:
\[
(x+1) - (x-1) = (2x-1) - (x+1)
\]
Simplifying both sides:
\[
2 = 2x - 2
\]
Solving for \( x \):
\[
2x = 4 \quad \Rightarrow \quad x = 2
\]

Thus, the value of \( x \) is 2.

---

Do you need more detailed steps for any of these problems?

### Related Questions:
1. How do you determine the common ratio in a geometric progression?
2. What is the general term for an arithmetic progression?
3. How can you prove relationships between consecutive terms in sequences?
4. What is the formula for the sum of a finite geometric progression?
5. How can you check if a sequence of numbers forms an arithmetic progression?

### Tip:
Always double-check whether the progression is arithmetic or geometric before applying formulas!

This page provides a detailed and concise solution to problems 26 to 30 involving arithmetic and geometric progressions, algebraic sequences, and series. Suitable for high school students, these problems cover essential formulas and concepts.

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