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

Question

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

UpStudy AI · Accepted Answer

To determine the first three terms and the $$ n^{\text{th}} $$ term of a sequence given its partial sum $$ S_n $$, we'll use the following approach:

1. **First Three Terms ($$ a_1, a_2, a_3 $$)**:
   - $$ a_1 = S_1 $$
   - $$ a_2 = S_2 - S_1 $$
   - $$ a_3 = S_3 - S_2 $$

2. **$$ n^{\text{th}} $$ Term ($$ a_n $$)**:
   - $$ a_n = S_n - S_{n-1} $$

Let's apply this method to both problems:

---

## Problem 3.1

**Given:**
$$ S_n = 43n - 6n^2 $$

### Finding the First Three Terms

1. **First Term ($$ a_1 $$):**
   $$
   a_1 = S_1 = 43(1) - 6(1)^2 = 43 - 6 = 37
   $$

2. **Second Term ($$ a_2 $$):**
   $$
   S_2 = 43(2) - 6(2)^2 = 86 - 24 = 62
   $$
   $$
   a_2 = S_2 - S_1 = 62 - 37 = 25
   $$

3. **Third Term ($$ a_3 $$):**
   $$
   S_3 = 43(3) - 6(3)^2 = 129 - 54 = 75
   $$
   $$
   a_3 = S_3 - S_2 = 75 - 62 = 13
   $$

**First Three Terms:**
$$
a_1 = 37,\quad a_2 = 25,\quad a_3 = 13
$$

### Finding the $$ n^{\text{th}} $$ Term

$$
a_n = S_n - S_{n-1} = [43n - 6n^2] - [43(n-1) - 6(n-1)^2]
$$
$$
= 43n - 6n^2 - 43n + 43 - 6(n^2 - 2n + 1)
$$
$$
= 43n - 6n^2 - 43n + 43 - 6n^2 + 12n - 6
$$
$$
= (-12n) + 37
$$
$$
\boxed{a_n = 49 - 12n}
$$

---

## Problem 3.3

**Given:**
$$ S_n = \frac{31}{2}n - \frac{3}{2}n^2 $$

### Finding the First Three Terms

1. **First Term ($$ a_1 $$):**
   $$
   a_1 = S_1 = \frac{31}{2}(1) - \frac{3}{2}(1)^2 = \frac{31}{2} - \frac{3}{2} = \frac{28}{2} = 14
   $$

2. **Second Term ($$ a_2 $$):**
   $$
   S_2 = \frac{31}{2}(2) - \frac{3}{2}(2)^2 = 31 - 6 = 25
   $$
   $$
   a_2 = S_2 - S_1 = 25 - 14 = 11
   $$

3. **Third Term ($$ a_3 $$):**
   $$
   S_3 = \frac{31}{2}(3) - \frac{3}{2}(3)^2 = \frac{93}{2} - \frac{27}{2} = \frac{66}{2} = 33
   $$
   $$
   a_3 = S_3 - S_2 = 33 - 25 = 8
   $$

**First Three Terms:**
$$
a_1 = 14,\quad a_2 = 11,\quad a_3 = 8
$$

### Finding the $$ n^{\text{th}} $$ Term

$$
a_n = S_n - S_{n-1} = \left(\frac{31}{2}n - \frac{3}{2}n^2\right) - \left(\frac{31}{2}(n-1) - \frac{3}{2}(n-1)^2\right)
$$
$$
= \frac{31}{2}n - \frac{3}{2}n^2 - \frac{31}{2}n + \frac{31}{2} - \frac{3}{2}(n^2 - 2n + 1)
$$
$$
= \frac{31}{2}n - \frac{3}{2}n^2 - \frac{31}{2}n + \frac{31}{2} - \frac{3}{2}n^2 + 3n - \frac{3}{2}
$$
$$
= ( -3n ) + 17
$$
$$
\boxed{a_n = 17 - 3n}
$$

---

## Summary

- **Problem 3.1:**
  - **First Three Terms:** $$ 37, \ 25, \ 13 $$
  - **$$ n^{\text{th}} $$ Term:** $$ a_n = 49 - 12n $$

- **Problem 3.3:**
  - **First Three Terms:** $$ 14, \ 11, \ 8 $$
  - **$$ n^{\text{th}} $$ Term:** $$ a_n = 17 - 3n $$
**Problem 3.1:** - **First Three Terms:** 37, 25, 13 - **$$ n^{\text{th}} $$ Term:** $$ a_n = 49 - 12n $$ **Problem 3.3:** - **First Three Terms:** 14, 11, 8 - **$$ n^{\text{th}} $$ Term:** $$ a_n = 17 - 3n $$

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

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