(11) The sum of the first three terms of an arithmetic sequence is 36 and their product is 1140 . Determine the three terms. (0) A certain sequence has a constant second difference of 8 . The sixth term of the sequence is 172 and the tenth term is 452 . Determine the general term. (p) The sum of the first $$ n $$ terms of a series is given by $$ S_{n}=\frac{n}{4}(7-2 n) $$. (1) Calculate the sum of the first 30 terms of this series.

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### **Problem (11):**

**_The sum of the first three terms of an arithmetic sequence is 36 and their product is 1140. Determine the three terms._**

**Solution:**

1. **Let the three terms be:**  
   $$ a - d $$, $$ a $$, $$ a + d $$  
   where $$ a $$ is the middle term and $$ d $$ is the common difference.

2. **Sum of the terms:**  
   $$
   (a - d) + a + (a + d) = 3a = 36 \implies a = 12
   $$

3. **Product of the terms:**  
   $$
   (a - d)(a)(a + d) = a(a^2 - d^2) = 1140
   $$  
   Substitute $$ a = 12 $$:  
   $$
   12(144 - d^2) = 1140 \implies 144 - d^2 = 95 \implies d^2 = 49 \implies d = \pm 7
   $$

4. **Determine the terms:**  
   - For $$ d = 7 $$: $$ 12 - 7 = 5 $$, $$ 12 $$, $$ 12 + 7 = 19 $$  
   - For $$ d = -7 $$: $$ 12 + 7 = 19 $$, $$ 12 $$, $$ 12 - 7 = 5 $$

**Answer:**  
The three terms are **5**, **12**, and **19**.

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### **Problem (0):**

**_A certain sequence has a constant second difference of 8. The sixth term of the sequence is 172 and the tenth term is 452. Determine the general term._**

**Solution:**

1. **Nature of the Sequence:**  
   A sequence with a constant second difference is quadratic. Its general term is:  
   $$
   a_n = an^2 + bn + c
   $$

2. **Given Constant Second Difference:**  
   The second difference for a quadratic sequence is $$ 2a = 8 $$, so $$ a = 4 $$.

3. **General Term So Far:**  
   $$
   a_n = 4n^2 + bn + c
   $$

4. **Using Given Terms to Set Up Equations:**
   - For the 6th term ($$ n = 6 $$):  
     $$
     4(6)^2 + 6b + c = 172 \implies 144 + 6b + c = 172 \implies 6b + c = 28 \quad \text{(Equation 1)}
     $$
   - For the 10th term ($$ n = 10 $$):  
     $$
     4(10)^2 + 10b + c = 452 \implies 400 + 10b + c = 452 \implies 10b + c = 52 \quad \text{(Equation 2)}
     $$

5. **Solving the Equations:**
   - Subtract Equation 1 from Equation 2:  
     $$
     (10b + c) - (6b + c) = 52 - 28 \implies 4b = 24 \implies b = 6
     $$
   - Substitute $$ b = 6 $$ into Equation 1:  
     $$
     6(6) + c = 28 \implies 36 + c = 28 \implies c = -8
     $$

5. **Final General Term:**  
   $$
   a_n = 4n^2 + 6n - 8
   $$

**Answer:**  
The general term is **$$ a_n = 4n² + 6n - 8 $$**.

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### **Problem (p) & (1):**

**_The sum of the first $$ n $$ terms of a series is given by $$ S_{n}=\frac{n}{4}(7-2n) $$. Calculate the sum of the first 30 terms of this series._**

**Solution:**

1. **Given Sum Formula:**  
   $$
   S_n = \frac{n}{4}(7 - 2n)
   $$

2. **Calculate $$ S_{30} $$:**  
   $$
   S_{30} = \frac{30}{4}(7 - 2 \times 30) = \frac{30}{4}(7 - 60) = \frac{30}{4} \times (-53)
   $$

3. **Simplify the Calculation:**  
   $$
   S_{30} = \frac{30 \times (-53)}{4} = \frac{-1590}{4} = -397.5 \quad \text{or} \quad -\frac{795}{2}
   $$

**Answer:**  
The sum of the first **30 terms** is **$$-397.5$$**.

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If you have any further questions or need additional assistance with these problems, feel free to ask!
**Problem (11):** The three terms are **5**, **12**, and **19**. **Problem (0):** The general term is **$$ a_n = 4n² + 6n - 8 $$**. **Problem (p) & (1):** The sum of the first **30 terms** is **-397.5**.

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