1.4 A quadratic pattern has a second term equal to 1 and a third term equal to -6 and a fifth term equal to -14 . 1.4 .1 calculate the second differences of this quadratic pattern 1.4 .2 hence, or otherwise, calculate the first term of the pattern 1.5 $$ \begin{array}{l} ext { The sequence } 4 ; 9 ext {; The pattern } 4 ; 9 ; x, 37 ; \ldots ext { is a quadratic pattern } \ 1.5 .1 ext { Calculate the value of } x ext {. } \ 1.5 .2 ext { (3) } \ ext { hence or otherwise determine the general term of this pattern. } \ 1.6\end{array} \begin{array}{l} ext { Consider the sequence: } 3 ; p ; 10 ; q ; 21 ext {. The sequence has a constant second } \ ext { difference of 1. Determine the values of } p ext { and } q ext {. }\end{array} $$

Question

1.4 A quadratic pattern has a second term equal to 1 and a third term equal to -6 and a fifth term equal to -14 . 1.4 .1 calculate the second differences of this quadratic pattern 1.4 .2 hence, or otherwise, calculate the first term of the pattern 1.5 $$ \begin{array}{l}	ext { The sequence } 4 ; 9 	ext {; The pattern } 4 ; 9 ; x, 37 ; \ldots 	ext { is a quadratic pattern } \ 1.5 .1 	ext { Calculate the value of } x 	ext {. } \ 1.5 .2 	ext { (3) } \ 	ext { hence or otherwise determine the general term of this pattern. } \ 1.6\end{array} \begin{array}{l}	ext { Consider the sequence: } 3 ; p ; 10 ; q ; 21 	ext {. The sequence has a constant second } \ 	ext { difference of 1. Determine the values of } p 	ext { and } q 	ext {. }\end{array} $$

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Let's break down the problem step by step.

### 1.4 Quadratic Pattern

We know that a quadratic pattern can be expressed in the form:
$$
a_n = An^2 + Bn + C
$$
where $$A$$, $$B$$, and $$C$$ are constants, and $$n$$ is the term number.

#### 1.4.1 Calculate the second differences

Given:
- Second term ($$a_2$$) = 1
- Third term ($$a_3$$) = -6
- Fifth term ($$a_5$$) = -14

We can express these terms using the quadratic formula:
- $$a_2 = 4A + 2B + C = 1$$
- $$a_3 = 9A + 3B + C = -6$$
- $$a_5 = 25A + 5B + C = -14$$

We can set up the following system of equations:

1. $$4A + 2B + C = 1$$  (Equation 1)
2. $$9A + 3B + C = -6$$ (Equation 2)
3. $$25A + 5B + C = -14$$ (Equation 3)

Next, we will solve this system of equations to find $$A$$, $$B$$, and $$C$$.

#### 1.4.2 Calculate the first term of the pattern

The first term $$a_1$$ can be calculated using the values of $$A$$, $$B$$, and $$C$$.

Let's solve the system of equations.

### Solving the System of Equations

We will subtract Equation 1 from Equation 2 and Equation 2 from Equation 3 to eliminate $$C$$.

1. Subtract Equation 1 from Equation 2:
$$
(9A + 3B + C) - (4A + 2B + C) = -6 - 1
$$
This simplifies to:
$$
5A + B = -7 \quad \text{(Equation 4)}
$$

2. Subtract Equation 2 from Equation 3:
$$
(25A + 5B + C) - (9A + 3B + C) = -14 + 6
$$
This simplifies to:
$$
16A + 2B = -8 \quad \text{(Equation 5)}
$$

Now we can solve Equations 4 and 5 together.

### Solving Equations 4 and 5

From Equation 4:
$$
B = -7 - 5A
$$

Substituting $$B$$ into Equation 5:
$$
16A + 2(-7 - 5A) = -8
$$
This simplifies to:
$$
16A - 14 - 10A = -8
$$
$$
6A - 14 = -8
$$
$$
6A = 6
$$
$$
A = 1
$$

Now substituting $$A$$ back into Equation 4 to find $$B$$:
$$
5(1) + B = -7
$$
$$
B = -12
$$

Now substituting $$A$$ and $$B$$ back into Equation 1 to find $$C$$:
$$
4(1) + 2(-12) + C = 1
$$
$$
4 - 24 + C = 1
$$
$$
C = 21
$$

Now we have:
- $$A = 1$$
- $$B = -12$$
- $$C = 21$$

The first term $$a_1$$ can be calculated as:
$$
a_1 = A(1^2) + B(1) + C = 1 - 12 + 21 = 10
$$

### 1.5 Quadratic Pattern

Given the sequence $$4, 9, x, 37, \ldots$$, we know that it is a quadratic pattern.

#### 1.5.1 Calculate the value of $$x$$

We can express the terms as:
- $$a_1 = 4$$
- $$a_2 = 9$$
- $$a_3 = x$$
- $$a_4 = 37$$

We can find the second differences to determine $$x$$.

The first differences are:
- $$d_1 = 9 - 4 = 5$$
- $$d_2 = x - 9$$
- $$d_3 = 37 - x$$

The second differences are:
- $$d_2 - d_1 = (x - 9) - 5 = x - 14$$
- $$d_3 - d_2 = (37 - x) - (x - 9) = 46 - 2x$$

Since the second differences are constant, we can set them equal:
$$
x - 14 = 46 - 2x
$$
Solving for $$x$$:
$$
3x = 60
$$
$$
x = 20
$$

#### 1.5.2 Determine the general term of this pattern

Now we have:
- $$a_1 = 4$$
- $$a_2 = 9$$
- $$a_3 = 20$$
- $$a_4 = 37$$

We can find the general term $$a_n = An^2 + Bn + C$$ using the known terms.

Using the values of $$A$$, $$B$$, and $$C$$ from the previous calculations, we can find the general term.

### 1.6 Sequence with Constant Second Difference

Given the sequence $$3, p, 10, q, 21$$ with a constant second difference of 1.

The first differences are:
- $$d_1 = p - 3$$
- $$d_2 = 10 - p$$
- $$d_3 = q - 10$$
- $$d_4 = 21 - q$$

The second differences are:
- $$d_2 - d_1 = (10 - p) - (p - 3) = 13 - 2p$$
- $$d_3 - d_2 = (q - 10) - (10 - p) = q - 20 + p$$
- $$d_4 - d_3 = (21 - q) - (q - 10) = 31 - 2q$$

Since the second differences are constant and equal to 1, we can set up the equations:
1. $$13 - 2p = 1$$
2. $$q - 20 + p = 1$$
3. $$31 - 2q = 1$$

Let's solve these equations to find $$p$$ and $$q$$.

### Solving for $$p$$ and $$q$$

1. From $$13
**1.4 Quadratic Pattern** - **Second Differences:** Calculated to be 1. - **First Term:** 10. **1.5 Quadratic Pattern** - **Value of $$x$$:** 20. - **General Term:** $$a_n = n^2 - 11n + 14$$. **1.6 Sequence with Constant Second Difference** - **Values of $$p$$ and $$q$$:** - $$p = 6$$ - $$q = 15$$

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