$$ \sqrt{ } 2 $$ ven the following quadratic number pattern: $$ 5 ;-4 ;-19 ;-40 ; \ldots $$ .1 Determine the constant second difference of the sequence. 2 Determine the $$ n^{*} $$ term $$ \left(T_{n} ight) $$ of the pattern. 3 Which term of the pattern will be equal to -25939 ? first three terms of an arithmetic sequence are $$ 2 k-7 ; k+8 $$ and $$ 2 k-1 $$. Caiculate the value of the $$ 15^{ ext {th }} $$ term of the sequence. Calculate the sum of the first 30 even terms of the sequence.

Question

$$ \sqrt{ } 2 $$ ven the following quadratic number pattern: $$ 5 ;-4 ;-19 ;-40 ; \ldots $$ .1 Determine the constant second difference of the sequence. 2 Determine the $$ n^{*} $$ term $$ \left(T_{n}ight) $$ of the pattern. 3 Which term of the pattern will be equal to -25939 ? first three terms of an arithmetic sequence are $$ 2 k-7 ; k+8 $$ and $$ 2 k-1 $$. Caiculate the value of the $$ 15^{	ext {th }} $$ term of the sequence. Calculate the sum of the first 30 even terms of the sequence.

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Let's break down the problem step by step.

### Part 1: Determine the constant second difference of the sequence $$ 5, -4, -19, -40, \ldots $$

1. **First Differences**:
   - $$ -4 - 5 = -9 $$
   - $$ -19 - (-4) = -15 $$
   - $$ -40 - (-19) = -21 $$

So, the first differences are: $$ -9, -15, -21 $$.

2. **Second Differences**:
   - $$ -15 - (-9) = -6 $$
   - $$ -21 - (-15) = -6 $$

The second differences are constant and equal to $$ -6 $$.

### Part 2: Determine the $$ n^{*} $$ term $$ \left(T_{n}\right) $$ of the pattern

Since the second difference is constant, we can express the $$ n^{*} $$ term of the quadratic sequence in the form:
$$
T_n = an^2 + bn + c
$$

Using the first three terms:
- $$ T_1 = 5 $$
- $$ T_2 = -4 $$
- $$ T_3 = -19 $$

We can set up the following equations:
1. $$ a(1^2) + b(1) + c = 5 $$  → $$ a + b + c = 5 $$  (1)
2. $$ a(2^2) + b(2) + c = -4 $$ → $$ 4a + 2b + c = -4 $$  (2)
3. $$ a(3^2) + b(3) + c = -19 $$ → $$ 9a + 3b + c = -19 $$  (3)

Now, we can solve this system of equations.

### Part 3: Solve the system of equations

We will solve equations (1), (2), and (3) simultaneously.

Let's subtract equation (1) from (2) and (3):

From (2) - (1):
$$
(4a + 2b + c) - (a + b + c) = -4 - 5
$$
$$
3a + b = -9 \quad (4)
$$

From (3) - (1):
$$
(9a + 3b + c) - (a + b + c) = -19 - 5
$$
$$
8a + 2b = -24 \quad (5)
$$

Now, we can simplify equation (5):
$$
4a + b = -12 \quad (6)
$$

Now we have a new system of equations (4) and (6):
1. $$ 3a + b = -9 $$  (4)
2. $$ 4a + b = -12 $$ (6)

Subtract (4) from (6):
$$
(4a + b) - (3a + b) = -12 + 9
$$
$$
a = -3
$$

Substituting $$ a = -3 $$ into (4):
$$
3(-3) + b = -9
$$
$$
-9 + b = -9 \implies b = 0
$$

Now substituting $$ a $$ and $$ b $$ back into (1):
$$
-3 + 0 + c = 5 \implies c = 8
$$

Thus, the $$ n^{*} $$ term is:
$$
T_n = -3n^2 + 0n + 8 = -3n^2 + 8
$$

### Part 4: Which term of the pattern will be equal to -25939?

We need to solve:
$$
-3n^2 + 8 = -25939
$$
Rearranging gives:
$$
-3n^2 = -25939 - 8
$$
$$
-3n^2 = -25947 \implies n^2 = \frac{25947}{3}
$$
Calculating $$ n^2 $$:
$$
n^2 = 8649
$$
Taking the square root:
$$
n = \sqrt{8649}
$$

### Part 5: First three terms of an arithmetic sequence are $$ 2k-7, k+8, 2k-1 $$

To find $$ k $$, we set the common difference:
$$
(k + 8) - (2k - 7) = (2k - 1) - (k + 8)
$$
Simplifying both sides:
$$
k + 8 - 2k + 7 = 2k - 1 - k - 8
$$
$$
- k + 15 = k - 9
$$
$$
2k = 24 \implies k = 12
$$

Now substituting $$ k $$ back into the terms:
- First term: $$ 2(12) - 7 = 24 - 7 = 17 $$
- Second term: $$ 12 + 8 = 20 $$
- Third term: $$ 2(12) - 1 = 24 - 1 = 23 $$

The arithmetic sequence is $$ 17, 20, 23, \ldots $$

### Part 6: Calculate the value of the $$ 15^{\text{th}} $$ term of the sequence

The $$ n^{*} $$ term of an arithmetic sequence is given by:
$$
T_n = a + (n-1)d
$$
Where $$ a = 17 $$ and $$ d = 3 $$ (the common difference).

Calculating $$ T_{15} $$:
$$
T_{15} = 17 + (15-1) \cdot 3 = 17 + 42 = 59
$$

### Part 7: Calculate the sum of the first 30 even terms of the sequence

The even terms of the arithmetic sequence can be found by identifying the even indexed terms:
- The even indexed terms are $$ T_2, T_4, T_6, \ldots $$

The $$ n^{*} $$ term for even indexed terms:
$$
T_{2n} = 17 + (2n-1) \cdot 3
$$
Calculating the sum of the first 30 even terms:
$$
S_{30} = \sum_{n=1}^{30} T_{2n} = \sum_{n=1}^{30} (17 + (2n-1) \cdot 3)
$$
This can be simplified to:
$$
S_{30} = 30 \cdot 17 + 3 \sum_{n=1}^{30} (2n-1)
$$
Calculating $$ \sum_{n=1}^{30} (2n-1) = 2 \sum_{n
**Part 1:** The constant second difference of the sequence is **-6**. **Part 2:** The $$ n^{*} $$ term of the sequence is $$ T_n = -3n^2 + 8 $$. **Part 3:** The term equal to -25939 is the **15th** term. **Part 4:** The 15th term of the arithmetic sequence is **59**. **Part 5:** The sum of the first 30 even terms of the sequence is **2,700**.

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