I see the uploaded image contains a set of math problems. I'll extract the content and provide step-by-step solutions to each question. Let me process and solve these for you.### Problem 1: The sequence is defined by $a_n = \frac{12n}{5 - (-1)^n}$. We need to find the sum of the first five terms.

Step 1: Calculate the terms.

• For $n = 1$: $a_1 = \frac{12 \cdot 1}{5 - (-1)^1} = \frac{12}{5 - (-1)} = \frac{12}{5 - (-1)} = \frac{12}{6} = 2$.
• For $n = 2$: $a_2 = \frac{12 \cdot 2}{5 - (-1)^2} = \frac{24}{5 - 1} = \frac{24}{4} = 6$.
• For $n = 3$: $a_3 = \frac{12 \cdot 3}{5 - (-1)^3} = \frac{36}{5 - (-1)} = \frac{36}{6} = 6$.
• For $n = 4$: $a_4 = \frac{12 \cdot 4}{5 - (-1)^4} = \frac{48}{5 - 1} = \frac{48}{4} = 12$.
• For $n = 5$: $a_5 = \frac{12 \cdot 5}{5 - (-1)^5} = \frac{60}{5 - (-1)} = \frac{60}{6} = 10$.

Step 2: Find the sum. $S_5 = a_1 + a_2 + a_3 + a_4 + a_5 = 2 + 6 + 6 + 12 + 10 = 36$.

Answer: $\text{A) } 36$.

Problem 2:

The sequence is defined as:

• $a_1 = -2$,
• $a_2 = 1$,
• $a_{n+2} = 3a_{n+1} + 2a_n$.

We need to find $a_5$.

Step 1: Calculate the terms.

• $a_3 = 3a_2 + 2a_1 = 3(1) + 2(-2) = 3 - 4 = -1$,
• $a_4 = 3a_3 + 2a_2 = 3(-1) + 2(1) = -3 + 2 = -1$,
• $a_5 = 3a_4 + 2a_3 = 3(-1) + 2(-1) = -3 - 2 = -5$.

Answer: $\text{A) } -5$.

Problem 3:

Which sequence is not increasing?

• $\text{A) } a_n = 2n - 3$ is linear with a positive slope ($a_n$ increases).
• $\text{B) } a_n = \left( \frac{1}{3} \right)^n$ is exponential decay ($a_n$ decreases, so it's not increasing).
• $\text{C) } a_n = 3^n + 3$ is exponential growth ($a_n$ increases).
• $\text{D) } a_n = (-3)^n + 1$ alternates, but its magnitude grows.

Answer: $\text{B) } a_n = \left( \frac{1}{3} \right)^n$.

Problem 4:

We need the 9th term of an arithmetic progression where:

• $a_1 = 5$,
• $d = -6$.

The general formula is $a_n = a_1 + (n-1)d$.

Step 1: Substitute. $a_9 = 5 + (9-1)(-6) = 5 + 8(-6) = 5 - 48 = -43$.

Answer: $\text{B) } -43$.

Problem 5:

We need the 3rd term of a geometric progression where:

• $a_1 = -3$,
• $q = -4$.

The general formula is $a_n = a_1 \cdot q^{n-1}$.

Step 1: Substitute. $a_3 = (-3) \cdot (-4)^{3-1} = (-3) \cdot (-4)^2 = (-3) \cdot 16 = -48$.

Answer: $\text{B) } -48$.

Let me know if you need detailed clarifications or further help!

Here are five related questions:

1. How do you identify arithmetic vs. geometric progressions?
2. Can you derive the formula for the sum of an arithmetic series?
3. How does alternating sequences like $(-3)^n$ behave?
4. What conditions make a sequence strictly increasing or decreasing?
5. What is the impact of the common ratio in geometric sequences?

Tip: Always verify the general formula of a sequence before calculating terms.