https://file.aiquickdraw.com/mathbot/file/1729577903689k29y9s4r.jpg

Let’s solve this step by step.

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### (a) **Arithmetic Progression (AP)**  
Given:
- Second term: \(a_2 = 7\)
- Thirteenth term: \(a_{13} = 10 \times a_1\)

The formula for the \(n\)-th term of an arithmetic progression is:  
\[
a_n = a_1 + (n - 1) \cdot d
\]  
where \(a_1\) is the first term and \(d\) is the common difference.

- For the second term:  
\[
a_2 = a_1 + (2 - 1) \cdot d = a_1 + d = 7
\]
Thus:
\[
a_1 + d = 7 \quad \text{(Equation 1)}
\]

- For the thirteenth term:  
\[
a_{13} = a_1 + (13 - 1) \cdot d = a_1 + 12d
\]
It is given that:
\[
a_{13} = 10 \times a_1
\]
Thus:
\[
a_1 + 12d = 10a_1 \quad \text{(Equation 2)}
\]
Rearrange Equation 2:
\[
12d = 10a_1 - a_1 = 9a_1 \quad \Rightarrow \quad d = \frac{9a_1}{12} = \frac{3a_1}{4}
\]

**Now solve for \(a_1\) and \(d\):**

Substitute \(d = \frac{3a_1}{4}\) into Equation 1:
\[
a_1 + \frac{3a_1}{4} = 7
\]
\[
\frac{4a_1 + 3a_1}{4} = 7 \quad \Rightarrow \quad \frac{7a_1}{4} = 7
\]
\[
7a_1 = 28 \quad \Rightarrow \quad a_1 = 4
\]

Now, substitute \(a_1 = 4\) into Equation 1:
\[
4 + d = 7 \quad \Rightarrow \quad d = 3
\]

Thus, the **first term** is \(a_1 = 4\) and the **common difference** is \(d = 3\).

---

### (b) **Geometric Series (GS)**  
Given:
- Second term: \(a_2 = 5\)
- Sum to infinity: \(S_\infty = 20\)

The formula for the sum to infinity of a geometric series is:
\[
S_\infty = \frac{a_1}{1 - r}
\]
where \(a_1\) is the first term and \(r\) is the common ratio (\(|r| < 1\)).

Also, the formula for the second term is:
\[
a_2 = a_1 \cdot r
\]
Thus:
\[
5 = a_1 \cdot r \quad \Rightarrow \quad a_1 = \frac{5}{r} \quad \text{(Equation 3)}
\]

Now, substitute \(a_1 = \frac{5}{r}\) into the formula for \(S_\infty\):
\[
20 = \frac{\frac{5}{r}}{1 - r}
\]
Simplify:
\[
20 = \frac{5}{r(1 - r)}
\]
\[
5 = 20r(1 - r)
\]
\[
5 = 20r - 20r^2 \quad \Rightarrow \quad 20r^2 - 20r + 5 = 0
\]
Divide by 5:
\[
4r^2 - 4r + 1 = 0
\]
This is a quadratic equation in \(r\). Solve using the quadratic formula:
\[
r = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(4)(1)}}{2(4)} = \frac{4 \pm \sqrt{16 - 16}}{8} = \frac{4 \pm 0}{8} = \frac{4}{8} = \frac{1}{2}
\]
Thus, \(r = \frac{1}{2}\).

Now, substitute \(r = \frac{1}{2}\) into Equation 3:
\[
a_1 = \frac{5}{\frac{1}{2}} = 5 \times 2 = 10
\]

Thus, the **first term** is \(a_1 = 10\) and the **common ratio** is \(r = \frac{1}{2}\).

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### Final Answers:
**(a)**  
- First term: \(a_1 = 4\)  
- Common difference: \(d = 3\)

**(b)**  
- First term: \(a_1 = 10\)  
- Common ratio: \(r = \frac{1}{2}\)

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Do you want further details or have any questions?  
Here are five follow-up questions to expand your understanding:

1. How do we derive the sum formula for an arithmetic progression?
2. What happens if the common ratio in a geometric series is greater than 1?
3. How do we determine the number of terms in a finite geometric series?
4. Can a geometric series with a negative common ratio have a sum to infinity?
5. What are the conditions for an arithmetic sequence to have a negative common difference?

**Tip:**  
When solving quadratic equations, always check if the discriminant is non-negative to ensure real solutions exist.

An arithmetic progression has the following properties: - its second term is 7; - its thirteenth term is ten times its first term. Determine the first term and the common difference. Also, find the geometric series that has the second term equal to 5 and sum to infinity equal to 20.

Learn how to solve problems involving arithmetic progression, including finding the first term and common difference, and geometric series, such as determining the sum to infinity with key mathematical formulas. Suitable for advanced high school math students.

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