QUESTION 2 KZN J16 Given the quadratic sequence: $$ 6: 6: 10: 18 $$; 2.1 Determine a formula for the $$ \mathrm{n}^{ ext {th }} $$ term of the sequence. 2.2 Determine between which two consecutive terms the first difference is 200? 2.3 Which term in the quadratic sequence has a value of 32010 ?

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QUESTION 2 KZN J16 Given the quadratic sequence: $$ 6: 6: 10: 18 $$; 2.1 Determine a formula for the $$ \mathrm{n}^{	ext {th }} $$ term of the sequence. 2.2 Determine between which two consecutive terms the first difference is 200? 2.3 Which term in the quadratic sequence has a value of 32010 ?

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**Step 1. Determine the $$ \mathrm{n}^{\text{th}} $$ term of the sequence**

We assume the quadratic sequence has a general term  
$$
a_n = an^2 + bn + c.
$$

Given the first three terms:  
$$
a_1 = 6,\quad a_2 = 6,\quad a_3 = 10.
$$

For $$ n=1 $$:
$$
a(1)^2 + b(1) + c = a + b + c = 6. \quad (1)
$$

For $$ n=2 $$:
$$
a(2)^2 + b(2) + c = 4a + 2b + c = 6. \quad (2)
$$

For $$ n=3 $$:
$$
a(3)^2 + b(3) + c = 9a + 3b + c = 10. \quad (3)
$$

Subtract equation (1) from equation (2):
$$
(4a + 2b + c) - (a + b + c) = 6 - 6 \quad \Longrightarrow \quad 3a + b = 0.
$$
Thus,
$$
b = -3a.
$$

Subtract equation (2) from equation (3):
$$
(9a + 3b + c) - (4a + 2b + c) = 10 - 6 \quad \Longrightarrow \quad 5a + b = 4.
$$
Substitute $$ b = -3a $$ into this:
$$
5a - 3a = 4 \quad \Longrightarrow \quad 2a = 4 \quad \Longrightarrow \quad a = 2.
$$
Now,  
$$
b = -3(2) = -6.
$$
Substitute $$ a $$ and $$ b $$ into equation (1):
$$
2 - 6 + c = 6 \quad \Longrightarrow \quad c = 10.
$$

Thus, the formula for the $$ \mathrm{n}^{\text{th}} $$ term is:
$$
a_n = 2n^2 - 6n + 10.
$$

---

**Step 2. Determine between which two consecutive terms the first difference is 200**

For a quadratic sequence, the first difference is given by:
$$
d_n = a_{n+1} - a_n.
$$
Using our formula:
$$
a_{n+1} = 2(n+1)^2 - 6(n+1) + 10.
$$
Expanding:
$$
\begin{aligned}
a_{n+1} &= 2(n^2 + 2n + 1) - 6n - 6 + 10 \
&= 2n^2 + 4n + 2 - 6n -6 +10 \
&= 2n^2 - 2n + 6.
\end{aligned}
$$
Subtract $$ a_n = 2n^2 - 6n + 10 $$:
$$
\begin{aligned}
d_n &= (2n^2 - 2n + 6) - (2n^2 -6n +10) \
&= 4n - 4 \
&= 4(n - 1).
\end{aligned}
$$
Set $$ d_n = 200 $$:
$$
4(n - 1) = 200 \quad \Longrightarrow \quad n - 1 = 50 \quad \Longrightarrow \quad n = 51.
$$
Thus, the first difference of 200 occurs between the 51st and 52nd term.

---

**Step 3. Determine which term in the sequence has a value of 32010**

We need to solve the equation:
$$
2n^2 - 6n + 10 = 32010.
$$
Subtract 32010 from both sides:
$$
2n^2 - 6n + 10 - 32010 = 0 \quad \Longrightarrow \quad 2n^2 - 6n - 32000 = 0.
$$
Divide the entire equation by 2:
$$
n^2 - 3n - 16000 = 0.
$$
Use the quadratic formula:
$$
n = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A},
$$
where $$ A = 1 $$, $$ B = -3 $$, and $$ C = -16000 $$. Then:
$$
\begin{aligned}
n &= \frac{3 \pm \sqrt{(-3)^2 - 4(1)(-16000)}}{2} \
&= \frac{3 \pm \sqrt{9 + 64000}}{2} \
&= \frac{3 \pm \sqrt{64009}}{2}.
\end{aligned}
$$
Note that
$$
253^2 = 64009.
$$
Thus,
$$
n = \frac{3 \pm 253}{2}.
$$
We obtain two solutions:
$$
n = \frac{256}{2} = 128 \quad \text{or} \quad n = \frac{-250}{2} = -125.
$$
Since $$ n $$ must be positive, we have:
$$
n = 128.
$$
Thus, the term with a value of 32010 is the 128th term.

---

**Final Answers:**

2.1 $$a_n = 2n^2 - 6n + 10$$  
2.2 The first difference of 200 is between the 51st and 52nd terms.  
2.3 The 128th term is 32010.
**Answers:** 2.1 The $$ \mathrm{n}^{\text{th}} $$ term is $$ 2n^2 - 6n + 10 $$. 2.2 The first difference of 200 occurs between the 51st and 52nd terms. 2.3 The 128th term is 32010.

QUESTION 2 KZN J16 Given the quadratic sequence: \( 6: 6: 10: 18 \); 2.1 Determine a formula for the \( \mathrm{n}^{\text {th }} \) term of the sequence. 2.2 Determine between which two consecutive terms the first difference is 200? 2.3 Which term in the quadratic sequence has a value of 32010 ?

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