3 Determine the first three terms and the $$ n^{ ext {th }} $$ term of the sequence if $$ 3.1 \mathrm{~S}_{n}=43 n-6 n^{2} $$ $$ 3.2 \mathrm{~S}_{n}=n+2 n^{2} $$ $$ 3.3 \quad \mathrm{~S}_{n}=\frac{31}{2} n-\frac{3}{2} n^{2} $$

Question

3 Determine the first three terms and the $$ n^{	ext {th }} $$ term of the sequence if $$ 3.1 \mathrm{~S}_{n}=43 n-6 n^{2} $$ $$ 3.2 \mathrm{~S}_{n}=n+2 n^{2} $$ $$ 3.3 \quad \mathrm{~S}_{n}=\frac{31}{2} n-\frac{3}{2} n^{2} $$

UpStudy AI · Accepted Answer

To determine the first three terms and the $$ n^{\text{th}} $$ term of each sequence given the partial sum $$ S_n $$, we'll use the relationship between the partial sums and the sequence terms. The $$ n^{\text{th}} $$ term $$ a_n $$ can be found using:

$$
a_n = S_n - S_{n-1}
$$

where $$ S_n $$ is the sum of the first $$ n $$ terms, and $$ S_{n-1} $$ is the sum of the first $$ n-1 $$ terms.

Let's solve each part step by step.

---

### **3.1 $$ S_n = 43n - 6n^2 $$**

**Finding $$ a_n $$:**
$$
a_n = S_n - S_{n-1} = (43n - 6n^2) - [43(n-1) - 6(n-1)^2]
$$
$$
= 43n - 6n^2 - 43n + 43 - 6(n^2 - 2n + 1)
$$
$$
= 43n - 6n^2 - 43n + 43 - 6n^2 + 12n - 6
$$
$$
= -12n + 37
$$

**Thus, the $$ n^{\text{th}} $$ term is:**
$$
a_n = -12n + 37
$$

**First three terms:**
$$
a_1 = -12(1) + 37 = 25 \
a_2 = -12(2) + 37 = 13 \
a_3 = -12(3) + 37 = 1
$$

*However, upon verifying with partial sums, there's a small correction:*

Using the correct formula:
$$
a_n = -12n + 49
$$
$$
a_1 = 37, \quad a_2 = 25, \quad a_3 = 13
$$

---

### **3.2 $$ S_n = n + 2n^2 $$**

**Finding $$ a_n $$:**
$$
a_n = S_n - S_{n-1} = (n + 2n^2) - [(n-1) + 2(n-1)^2]
$$
$$
= n + 2n^2 - n + 1 - 2(n^2 - 2n + 1)
$$
$$
= 2n^2 + 0n + 1 - 2n^2 + 4n - 2
$$
$$
= 4n - 1
$$

**Thus, the $$ n^{\text{th}} $$ term is:**
$$
a_n = 4n - 1
$$

**First three terms:**
$$
a_1 = 3, \quad a_2 = 7, \quad a_3 = 11
$$

---

### **3.3 $$ S_n = \frac{31}{2}n - \frac{3}{2}n^2 $$**

**Finding $$ a_n $$:**
$$
a_n = S_n - S_{n-1} = \left(\frac{31}{2}n - \frac{3}{2}n^2\right) - \left[\frac{31}{2}(n-1) - \frac{3}{2}(n-1)^2\right]
$$
$$
= \frac{31}{2}n - \frac{3}{2}n^2 - \frac{31}{2}n + \frac{31}{2} - \frac{3}{2}(n^2 - 2n + 1)
$$
$$
= -3n + 17
$$

**Thus, the $$ n^{\text{th}} $$ term is:**
$$
a_n = -3n + 17
$$

**First three terms:**
$$
a_1 = 14, \quad a_2 = 11, \quad a_3 = 8
$$

---

### **Summary**

1. **For $$ S_n = 43n - 6n^2 $$:**
   - **$$ n^{\text{th}} $$ term:** $$ a_n = -12n + 37 $$
   - **First three terms:** 37, 25, 13

2. **For $$ S_n = n + 2n^2 $$:**
   - **$$ n^{\text{th}} $$ term:** $$ a_n = 4n - 1 $$
   - **First three terms:** 3, 7, 11

3. **For $$ S_n = \frac{31}{2}n - \frac{3}{2}n^2 $$:**
   - **$$ n^{\text{th}} $$ term:** $$ a_n = -3n + 17 $$
   - **First three terms:** 14, 11, 8
**First Three Terms and $$ n^{\text{th}} $$ Term of Each Sequence** 1. **For $$ S_n = 43n - 6n^2 $$:** - **$$ n^{\text{th}} $$ Term:** $$ a_n = -12n + 37 $$ - **First Three Terms:** 37, 25, 13 2. **For $$ S_n = n + 2n^2 $$:** - **$$ n^{\text{th}} $$ Term:** $$ a_n = 4n - 1 $$ - **First Three Terms:** 3, 7, 11 3. **For $$ S_n = \frac{31}{2}n - \frac{3}{2}n^2 $$:** - **$$ n^{\text{th}} $$ Term:** $$ a_n = -3n + 17 $$ - **First Three Terms:** 14, 11, 8

3 Determine the first three terms and the \( n^{\text {th }} \) term of the sequence if \( 3.1 \mathrm{~S}_{n}=43 n-6 n^{2} \) \( 3.2 \mathrm{~S}_{n}=n+2 n^{2} \) \( 3.3 \quad \mathrm{~S}_{n}=\frac{31}{2} n-\frac{3}{2} n^{2} \)

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