QUESTION 3 3.1 $$ 5 ; 2 $$ and -1 are the first three terms of a linear pattern. 3.1.1 Calculate $$ T_{4} $$. (2) 3.1.2 Write down the general rule of the pattern. (2) 3.2 Given the quadratic pattern: $$ 12 ; 45 ; 100 ; 177 \ldots $$ 3.2.1 Write down the next term of the pattern. 3.2.2 Determine a formula for the $$ n^{ ext {th }} $$ term of the pattern. (2) (4) [10]

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QUESTION 3 3.1 $$ 5 ; 2 $$ and -1 are the first three terms of a linear pattern. 3.1.1 Calculate $$ T_{4} $$. (2) 3.1.2 Write down the general rule of the pattern. (2) 3.2 Given the quadratic pattern: $$ 12 ; 45 ; 100 ; 177 \ldots $$ 3.2.1 Write down the next term of the pattern. 3.2.2 Determine a formula for the $$ n^{	ext {th }} $$ term of the pattern. (2) (4) [10]

UpStudy AI · Accepted Answer

Let's solve the problem step by step.

### 3.1 Linear Pattern

The first three terms of the linear pattern are $$ 5, 2, -1 $$.

#### 3.1.1 Calculate $$ T_{4} $$

To find $$ T_{4} $$, we first need to determine the common difference $$ d $$ of the linear pattern. The common difference can be calculated as follows:

$$
d = T_{2} - T_{1} = 2 - 5 = -3
$$

Now, we can find $$ T_{3} $$ to confirm the pattern:

$$
T_{3} = T_{2} + d = 2 - 3 = -1
$$

Now, we can calculate $$ T_{4} $$:

$$
T_{4} = T_{3} + d = -1 - 3 = -4
$$

Thus, $$ T_{4} = -4 $$.

#### 3.1.2 Write down the general rule of the pattern

The general rule for a linear pattern can be expressed as:

$$
T_n = T_1 + (n-1) \cdot d
$$

Substituting $$ T_1 = 5 $$ and $$ d = -3 $$:

$$
T_n = 5 + (n-1)(-3) = 5 - 3(n-1) = 8 - 3n
$$

### 3.2 Quadratic Pattern

The given quadratic pattern is $$ 12, 45, 100, 177 $$.

#### 3.2.1 Write down the next term of the pattern

To find the next term, we first need to determine the differences between the terms:

1. First differences:
   - $$ 45 - 12 = 33 $$
   - $$ 100 - 45 = 55 $$
   - $$ 177 - 100 = 77 $$

So, the first differences are $$ 33, 55, 77 $$.

2. Second differences:
   - $$ 55 - 33 = 22 $$
   - $$ 77 - 55 = 22 $$

The second differences are constant at $$ 22 $$, indicating a quadratic pattern.

To find the next term, we can add the last first difference $$ 77 $$ to the last term $$ 177 $$:

$$
\text{Next first difference} = 77 + 22 = 99
$$
$$
\text{Next term} = 177 + 99 = 276
$$

Thus, the next term is $$ 276 $$.

#### 3.2.2 Determine a formula for the $$ n^{\text{th}} $$ term of the pattern

Since the second differences are constant, we can express the $$ n^{\text{th}} $$ term of a quadratic sequence in the form:

$$
T_n = an^2 + bn + c
$$

We can set up a system of equations using the first three terms:

1. For $$ n = 1 $$:
   $$
   a(1^2) + b(1) + c = 12 \quad \Rightarrow \quad a + b + c = 12 \quad (1)
   $$

2. For $$ n = 2 $$:
   $$
   a(2^2) + b(2) + c = 45 \quad \Rightarrow \quad 4a + 2b + c = 45 \quad (2)
   $$

3. For $$ n = 3 $$:
   $$
   a(3^2) + b(3) + c = 100 \quad \Rightarrow \quad 9a + 3b + c = 100 \quad (3)
   $$

Now, we can solve this system of equations.

Let's solve the equations (1), (2), and (3) together.

### Solving the System of Equations

We will use the tool to solve the system of equations:

$$
\begin{align*}
a + b + c &= 12 \
4a + 2b + c &= 45 \
9a + 3b + c &= 100
\end{align*}
$$

Now, I will proceed to solve this system.
Solve the system of equations by following steps:
- step0: Solve using the substitution method:
  $$\left\{ \begin{array}{l}a+b+c=12\4a+2b+c=45\9a+3b+c=100\end{array}\right.$$
- step1: Solve the equation:
  $$\left\{ \begin{array}{l}a=12-b-c\4a+2b+c=45\9a+3b+c=100\end{array}\right.$$
- step2: Substitute the value of $$a:$$
  $$\left\{ \begin{array}{l}4\left(12-b-c\right)+2b+c=45\9\left(12-b-c\right)+3b+c=100\end{array}\right.$$
- step3: Simplify:
  $$\left\{ \begin{array}{l}48-2b-3c=45\108-6b-8c=100\end{array}\right.$$
- step4: Solve the equation:
  $$\left\{ \begin{array}{l}b=\frac{3-3c}{2}\108-6b-8c=100\end{array}\right.$$
- step5: Substitute the value of $$b:$$
  $$108-6\times \frac{3-3c}{2}-8c=100$$
- step6: Simplify:
  $$99+c=100$$
- step7: Move the constant to the right side:
  $$c=100-99$$
- step8: Subtract the numbers:
  $$c=1$$
- step9: Substitute the value of $$c:$$
  $$b=\frac{3-3\times 1}{2}$$
- step10: Substitute back:
  $$b=\frac{3-3}{2}$$
- step11: Calculate:
  $$b=0$$
- step12: Substitute the values of $$b,c:$$
  $$a=12-0-1$$
- step13: Substitute back:
  $$a=12-1$$
- step14: Calculate:
  $$a=11$$
- step15: Calculate:
  $$\left\{ \begin{array}{l}a=11\b=0\c=1\end{array}\right.$$
- step16: Check the solution:
  $$\left\{ \begin{array}{l}a=11\b=0\c=1\end{array}\right.$$
- step17: Rewrite:
  $$\left(a,b,c\right) = \left(11,0,1\right)$$
The solution to the system of equations is:

$$
(a, b, c) = (11, 0, 1)
$$

Thus, the formula for the $$ n^{\text{th}} $$ term of the quadratic pattern is:

$$
T_n = 11n^2 + 0n + 1 = 11n^2 + 1
$$

### Summary of Answers

1. **3.1.1**: $$ T_{4} = -4 $$
2. **3.1.2**: General rule: $$ T_n = 8 - 3n $$
3. **3.2.1**: Next term: $$ 276 $$
4. **3.2.2**: Formula for the $$ n^{\text{th}} $$ term: $$ T_n = 11n^2 + 1 $$

This concludes the solution to the problem.
**3.1 Linear Pattern** 1. **Calculate $$ T_{4} $$:** $$ T_{4} = -4 $$ 2. **General Rule:** $$ T_n = 8 - 3n $$ **3.2 Quadratic Pattern** 1. **Next Term:** $$ 276 $$ 2. **Formula for the $$ n^{\text{th}} $$ Term:** $$ T_n = 11n^2 + 1 $$

QUESTION 3
3.1 and -1 are the first three terms of a linear pattern.
3.1.1 Calculate .
(2)
3.1.2 Write down the general rule of the pattern.
(2)
3.2 Given the quadratic pattern:
3.2.1 Write down the next term of the pattern.
3.2.2 Determine a formula for the term of the pattern.
(2)
(4)
[10]

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QUESTION 3 3.1 and -1 are the first three terms of a linear pattern. 3.1.1 Calculate . (2) 3.1.2 Write down the general rule of the pattern. (2) 3.2 Given the quadratic pattern: 3.2.1 Write down the next term of the pattern. 3.2.2 Determine a formula for the term of the pattern. (2) (4) [10]