ESTION 2 KZN J16 Determine a formula for the $$ \mathrm{n}^{ ext {a }} $$ term of the sequence Determine between which two consecutive terms the first difference, is 20 ? $$ 6: 10: \ldots $$ Which term in the quadratic sequence has a value of 32010 ?

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ESTION 2 KZN J16 Determine a formula for the $$ \mathrm{n}^{	ext {a }} $$ term of the sequence Determine between which two consecutive terms the first difference, is 20 ? $$ 6: 10: \ldots $$ Which term in the quadratic sequence has a value of 32010 ?

UpStudy AI · Accepted Answer

To solve the problem, we will follow these steps:

1. **Identify the sequence and find the formula for the $$ n^{\text{th}} $$ term.**
2. **Calculate the first difference and determine between which two consecutive terms it equals 20.**
3. **Find which term in the sequence has a value of 32010.**

### Step 1: Identify the Sequence and Find the Formula

The given sequence is $$ 6, 10, \ldots $$. This appears to be a quadratic sequence. To find the $$ n^{\text{th}} $$ term, we will first calculate the first and second differences.

- **First differences:**
  - $$ 10 - 6 = 4 $$

Since we only have two terms, we need to assume a quadratic form for the $$ n^{\text{th}} $$ term, which can be expressed as:
$$
a_n = An^2 + Bn + C
$$

We will use the known terms to set up equations.

For $$ n = 1 $$:
$$
A(1^2) + B(1) + C = 6 \quad \text{(1)}
$$

For $$ n = 2 $$:
$$
A(2^2) + B(2) + C = 10 \quad \text{(2)}
$$

### Step 2: Set Up the System of Equations

From the equations (1) and (2), we have:
1. $$ A + B + C = 6 $$
2. $$ 4A + 2B + C = 10 $$

Now, we can subtract equation (1) from equation (2):
$$
(4A + 2B + C) - (A + B + C) = 10 - 6
$$
This simplifies to:
$$
3A + B = 4 \quad \text{(3)}
$$

### Step 3: Find the Second Difference

Since we suspect this is a quadratic sequence, we need to find the second difference. The first difference is constant, so we can assume a quadratic form.

Let's assume the next term $$ a_3 $$ is $$ x $$. Then:
- First difference from $$ a_2 $$ to $$ a_3 $$ is $$ x - 10 $$.

The second difference should be constant. Let's denote the second difference as $$ d $$:
$$
d = (x - 10) - 4
$$

### Step 4: Solve for $$ x $$

We can set $$ d = 0 $$ for a quadratic sequence:
$$
(x - 10) - 4 = 0 \implies x - 14 = 0 \implies x = 14
$$

Now we have the first three terms: $$ 6, 10, 14 $$.

### Step 5: Find the General Formula

Now we can find the general formula. We have:
- $$ a_1 = 6 $$
- $$ a_2 = 10 $$
- $$ a_3 = 14 $$

We can see that the first differences are $$ 4, 4, \ldots $$, which suggests a linear increase.

The general formula for the $$ n^{\text{th}} $$ term can be expressed as:
$$
a_n = 6 + (n-1) \cdot 4 = 4n + 2
$$

### Step 6: Determine Between Which Two Consecutive Terms the First Difference is 20

To find when the first difference is 20, we set:
$$
a_{n+1} - a_n = 20
$$
Substituting the formula:
$$
(4(n+1) + 2) - (4n + 2) = 20
$$
This simplifies to:
$$
4 = 20 \quad \text{(which is not possible)}
$$

Since the first difference is constant at 4, it will never equal 20.

### Step 7: Find Which Term Has a Value of 32010

To find which term equals 32010, we set:
$$
4n + 2 = 32010
$$
Solving for $$ n $$:
$$
4n = 32008 \implies n = \frac{32008}{4} = 8002
$$

### Summary of Results

1. The formula for the $$ n^{\text{th}} $$ term is $$ a_n = 4n + 2 $$.
2. The first difference is always 4, so it never equals 20.
3. The term that has a value of 32010 is the $$ 8002^{\text{nd}} $$ term.
The $$ n^{\text{th}} $$ term of the sequence is $$ a_n = 4n + 2 $$. The first difference is always 4, so it never equals 20. The term with a value of 32010 is the 8002nd term.

ESTION 2 KZN J16 Determine a formula for the \( \mathrm{n}^{\text {a }} \) term of the sequence Determine between which two consecutive terms the first difference, is 20 ? \( 6: 10: \ldots \) Which term in the quadratic sequence has a value of 32010 ?

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