(n) The sum of the first three terms of an arithmetic sequence is 36 and their product is 1140 . Deternine the three terms. (o) 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 gencral 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 (n)**

**Given:**
- The sum of the first three terms of an arithmetic sequence is 36.
- The product of these three terms is 1140.

**Objective:** Determine the three terms.

**Solution:**

1. **Representation of Terms:**
   
   Let the three consecutive terms of the arithmetic sequence be:
   $$
   a, \quad a + d, \quad a + 2d
   $$
   where:
   - $$ a $$ = first term
   - $$ d $$ = common difference
   
2. **Sum of the Terms:**
   
   $$
   a + (a + d) + (a + 2d) = 3a + 3d = 36 \
   \Rightarrow a + d = 12 \quad \text{(Equation 1)}
   $$
   
3. **Product of the Terms:**
   
   $$
   a \times (a + d) \times (a + 2d) = 1140
   $$
   
   Substitute $$ d = 12 - a $$ from Equation 1:
   
   $$
   a \times 12 \times (24 - a) = 1140 \
   \Rightarrow a(24 - a) = \frac{1140}{12} = 95 \
   \Rightarrow a^2 - 24a + 95 = 0
   $$
   
4. **Solving the Quadratic Equation:**
   
   $$
   a = \frac{24 \pm \sqrt{(24)^2 - 4 \times 1 \times 95}}{2} = \frac{24 \pm \sqrt{576 - 380}}{2} = \frac{24 \pm \sqrt{196}}{2}
   $$
   
   $$
   a = \frac{24 \pm 14}{2}
   $$
   
   Thus, two possible values for $$ a $$:
   - $$ a = \frac{38}{2} = 19 $$
   - $$ a = \frac{10}{2} = 5 $$
   
5. **Determining the Terms:**
   
   - **If $$ a = 19 $$:**
     $$
     d = 12 - 19 = -7 \
     \text{Terms: } 19, \quad 12, \quad 5
     $$
   
   - **If $$ a = 5 $$:**
     $$
     d = 12 - 5 = 7 \
     \text{Terms: } 5, \quad 12, \quad 19
     $$
   
   Both sequences are valid arithmetic sequences with the given sum and product.

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

---

### **Problem (o)**

**Given:**
- A sequence has a constant second difference of 8.
- The sixth term ($$ a_6 $$) is 172.
- The tenth term ($$ a_{10} $$) is 452.

**Objective:** Determine the general term of the sequence.

**Solution:**

1. **Understanding Second Differences:**
   
   A constant second difference implies that the sequence is quadratic. The general form of a quadratic sequence is:
   $$
   a_n = An^2 + Bn + C
   $$
   
   The second difference ($$ \Delta^2 $$) of a quadratic sequence is $$ 2A $$. Given $$ \Delta^2 = 8 $$:
   $$
   2A = 8 \quad \Rightarrow \quad A = 4
   $$
   
   So, the general term becomes:
   $$
   a_n = 4n^2 + Bn + C
   $$
   
2. **Setting Up Equations:**
   
   Use the given terms to create equations:
   
   - For $$ n = 6 $$:
     $$
     4(6)^2 + B(6) + C = 172 \
     144 + 6B + C = 172 \
     6B + C = 28 \quad \text{(Equation 1)}
     $$
   
   - For $$ n = 10 $$:
     $$
     4(10)^2 + B(10) + C = 452 \
     400 + 10B + C = 452 \
     10B + C = 52 \quad \text{(Equation 2)}
     $$
   
3. **Solving the System of Equations:**
   
   Subtract Equation 1 from Equation 2:
   $$
   (10B + C) - (6B + C) = 52 - 28 \
   4B = 24 \
   B = 6
   $$
   
   Substitute $$ B = 6 $$ into Equation 1:
   $$
   6(6) + C = 28 \
   36 + C = 28 \
   C = -8
   $$
   
4. **Final General Term:**
   
   $$
   a_n = 4n^2 + 6n - 8
   $$
   
   **Verification:**
   
   - For $$ n = 6 $$:
     $$
     4(36) + 6(6) - 8 = 144 + 36 - 8 = 172 \quad \text{(Correct)}
     $$
   
   - For $$ n = 10 $$:
     $$
     4(100) + 6(10) - 8 = 400 + 60 - 8 = 452 \quad \text{(Correct)}
     $$

**Answer:**
The general term is $$ \mathbf{a_n = 4n^{2} + 6n - 8} $$.

---

### **Problem (p)**

**Given:**
- The sum of the first $$ n $$ terms of a series is:
  $$
  S_n = \frac{n}{4}(7 - 2n)
  $$
- Calculate the sum of the first 30 terms ($$ S_{30} $$).

**Solution:**

1. **Substitute $$ n = 30 $$ into the sum formula:**
   $$
   S_{30} = \frac{30}{4}(7 - 2 \times 30) = \frac{30}{4}(7 - 60) = \frac{30}{4}(-53)
   $$
   
2. **Simplify the Expression:**
   $$
   S_{30} = \frac{30}{4} \times (-53) = \frac{15}{2} \times (-53) = \frac{-795}{2} = -397.5
   $$
   
   Alternatively, it can be expressed as a fraction:
   $$
   S_{30} = -\frac{795}{2}
   $$

**Answer:**
The sum of the first 30 terms is **–795⁄2** (that is, $$ S_{30} = -\frac{795}{2} $$).
**Problem (n):** The three terms are **5, 12, and 19**. **Problem (o):** The general term is $$ \mathbf{a_n = 4n^{2} + 6n - 8} $$. **Problem (p):** The sum of the first 30 terms is **–795⁄2**.

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