I'll process the mathematical content from the uploaded images and provide solutions. Let me analyze them for clarity.The questions in the uploaded images are related to sequences, patterns, and algebraic simplifications. I'll address the key parts:

Image 1 Analysis

a) Consider the pattern: $4, -1, -6, -11, \ldots, -491$

1. Expression for the $n$-th term of the pattern:

   • This is an arithmetic sequence where the first term ($a$) is $4$, and the common difference ($d$) is $-1 - 4 = -5$.
   • Formula for the $n$-th term of an arithmetic sequence: $T_n = a + (n-1)d$ Substituting: $T_n = 4 + (n-1)(-5) = 4 - 5n + 5 = 9 - 5n$

2. Number of terms in the sequence:

   • Last term $T_n = -491$, so: $-491 = 9 - 5n$ Solve for $n$: $-491 - 9 = -5n \implies -500 = -5n \implies n = 100$ There are $100$ terms in the sequence.

3. Value of the 20th odd-number term in the sequence:

   • The odd terms in the sequence start at $4$, skipping every second term. Hence: $T_{2k-1} = 9 - 5(2k-1) = 9 - 10k + 5 = 14 - 10k$ For $k = 20$: $T_{39} = 14 - 10(20) = 14 - 200 = -186$

b) Sequence $x-4, 2x-1, 3x+2, \ldots$

1. First difference in terms of $x$:

   • Difference between consecutive terms: $D_1 = (2x - 1) - (x - 4) = 2x - 1 - x + 4 = x + 3$

2. If $T_n = xn + 3n - 7$ and $T_{20} = 153$, determine $x$:

   • Substituting $n = 20$: $T_{20} = 153 = x(20) + 3(20) - 7$ Simplify: $153 = 20x + 60 - 7 \implies 153 = 20x + 53 \implies 20x = 100 \implies x = 5$

Image 2 Analysis

a) Factorize expressions

1. $x^3 + y^3$:

   • This is a sum of cubes: $x^3 + y^3 = (x+y)(x^2 - xy + y^2)$

2. $ax^2 - bx^2 - 4a + 4b$:

   • Group terms: $ax^2 - bx^2 - 4a + 4b = x^2(a-b) - 4(a-b)$ Factorize: $(x^2 - 4)(a-b)$

b) Simplify expressions completely

1. $\frac{3^n + 3^{n+2}}{3^{n-1}}$:

   • Factorize numerator: $\frac{3^n(1 + 3^2)}{3^{n-1}} = \frac{3^n(10)}{3^{n-1}}$ Simplify: $10 \cdot 3 = 30$

2. $\frac{25^{2n-1}}{5^{3n+1} \cdot 5^{n-3}}$:

   • Rewrite bases: $25^{2n-1} = (5^2)^{2n-1} = 5^{4n-2}$ Combine exponents: $\frac{5^{4n-2}}{5^{3n+1} \cdot 5^{n-3}} = \frac{5^{4n-2}}{5^{3n+1+n-3}} = 5^{4n-2 - (4n-2)} = 1$

Would you like detailed explanations for any specific part? Here are 5 related questions:

1. How can you derive expressions for geometric sequences?
2. What is the general approach to simplify exponential expressions?
3. Can you explain how to factorize polynomials with four terms?
4. How do you determine a specific $k$-th term in any sequence?
5. What other methods exist to solve sequence-related problems?

Tip: Always check for patterns like constant differences or ratios when analyzing sequences.