[ION 2 $$ 7 ; x ; y ;-11 ; \ldots $$ is an arithmetic sequence. Determine the values of $$ x $$ and $$ y $$. (4) Given the quadratic number pattern: $$ -3 ; 6 ; 27 ; 60 ; \ldots $$ 2.2.1 Determine the general term of the pattern in the form $$ T_{n}=a n^{2}+b n+c $$. 2.2.2 Calculate the value of the $$ 50^{ ext {th }} $$ term of the pattern. 2.2.3 Show that the sum of the first $$ n $$ first-differences of this pattern can be given by $$ S_{n}=6 n^{2}+3 n $$. 2.2.4 How many consecutive first-differences were added to the first term of the quadratic number pattern to obtain a term in the quadratic number pattern that has a value of 21060 ? [17 ION 3 Prove that $$ \sum_{k=1}^{\infty} 4.3^{2-k} $$ is a convergent geometric series. Show ALL your calculations. If $$ \sum_{k=p}^{\infty} 4.3^{2-k}=\frac{2}{9} $$, determine the value of $$ p $$. (3

Question

[ION 2 $$ 7 ; x ; y ;-11 ; \ldots $$ is an arithmetic sequence. Determine the values of $$ x $$ and $$ y $$. (4) Given the quadratic number pattern: $$ -3 ; 6 ; 27 ; 60 ; \ldots $$ 2.2.1 Determine the general term of the pattern in the form $$ T_{n}=a n^{2}+b n+c $$. 2.2.2 Calculate the value of the $$ 50^{	ext {th }} $$ term of the pattern. 2.2.3 Show that the sum of the first $$ n $$ first-differences of this pattern can be given by $$ S_{n}=6 n^{2}+3 n $$. 2.2.4 How many consecutive first-differences were added to the first term of the quadratic number pattern to obtain a term in the quadratic number pattern that has a value of 21060 ? [17 ION 3 Prove that $$ \sum_{k=1}^{\infty} 4.3^{2-k} $$ is a convergent geometric series. Show ALL your calculations. If $$ \sum_{k=p}^{\infty} 4.3^{2-k}=\frac{2}{9} $$, determine the value of $$ p $$. (3

UpStudy AI · Accepted Answer

Sure, let's tackle each of these problems step by step.

---

## **ION 2**

### **Problem 7**

**Given:** An arithmetic sequence $$ 7 ; x ; y ; -11 ; \ldots $$

**Find:** The values of $$ x $$ and $$ y $$.

**Solution:**

In an arithmetic sequence, the difference between consecutive terms is constant. Let's denote this common difference as $$ d $$.

Given the sequence:
- First term ($$ a_1 $$) = 7
- Second term ($$ a_2 $$) = $$ x $$
- Third term ($$ a_3 $$) = $$ y $$
- Fourth term ($$ a_4 $$) = -11

Using the property of arithmetic sequences:
$$
a_{n} = a_{1} + (n-1)d
$$

Let's find the common difference $$ d $$ using the first and fourth terms.

$$
a_4 = a_1 + 3d \
-11 = 7 + 3d \
-11 - 7 = 3d \
-18 = 3d \
d = -6
$$

Now, find $$ x $$ and $$ y $$:

$$
a_2 = a_1 + d = 7 + (-6) = 1 \
a_3 = a_2 + d = 1 + (-6) = -5
$$

**Answer:** 
- $$ x = 1 $$
- $$ y = -5 $$

---

## **Quadratic Number Pattern**

**Given Pattern:** $$ -3 ; 6 ; 27 ; 60 ; \ldots $$

### **2.2.1 Determine the general term $$ T_{n} = a n^{2} + b n + c $$.**

**Solution:**

Assume the general term is quadratic:
$$
T_n = an^2 + bn + c
$$

We can set up equations using the given terms:

1. For $$ n = 1 $$:
$$
T_1 = a(1)^2 + b(1) + c = a + b + c = -3 \quad \text{(Equation 1)}
$$

2. For $$ n = 2 $$:
$$
T_2 = a(2)^2 + b(2) + c = 4a + 2b + c = 6 \quad \text{(Equation 2)}
$$

3. For $$ n = 3 $$:
$$
T_3 = a(3)^2 + b(3) + c = 9a + 3b + c = 27 \quad \text{(Equation 3)}
$$

Now, solve the system of equations.

**Subtract Equation 1 from Equation 2:**
$$
(4a + 2b + c) - (a + b + c) = 6 - (-3) \
3a + b = 9 \quad \text{(Equation 4)}
$$

**Subtract Equation 2 from Equation 3:**
$$
(9a + 3b + c) - (4a + 2b + c) = 27 - 6 \
5a + b = 21 \quad \text{(Equation 5)}
$$

**Subtract Equation 4 from Equation 5:**
$$
(5a + b) - (3a + b) = 21 - 9 \
2a = 12 \
a = 6
$$

**Substitute $$ a = 6 $$ into Equation 4:**
$$
3(6) + b = 9 \
18 + b = 9 \
b = -9
$$

**Substitute $$ a = 6 $$ and $$ b = -9 $$ into Equation 1:**
$$
6 - 9 + c = -3 \
-3 + c = -3 \
c = 0
$$

**Answer:**
$$
T_n = 6n^2 - 9n
$$

---

### **2.2.2 Calculate the value of the $$ 50^{\text{th}} $$ term of the pattern.**

**Solution:**

Using the general term $$ T_n = 6n^2 - 9n $$:

$$
T_{50} = 6(50)^2 - 9(50) = 6(2500) - 450 = 15000 - 450 = 14550
$$

**Answer:** $$ T_{50} = 14,\!550 $$

---

### **2.2.3 Show that the sum of the first $$ n $$ first-differences of this pattern can be given by $$ S_{n} = 6n^{2} + 3n $$.**

**Solution:**

First, find the first difference sequence. The first difference $$ D_n = T_n - T_{n-1} $$.

Given $$ T_n = 6n^2 - 9n $$:

$$
D_n = T_n - T_{n-1} = [6n^2 - 9n] - [6(n-1)^2 - 9(n-1)]
$$
$$
D_n = 6n^2 - 9n - [6(n^2 - 2n + 1) - 9n + 9]
$$
$$
D_n = 6n^2 - 9n - 6n^2 + 12n - 6 + 9n - 9
$$
$$
D_n = (6n^2 - 6n^2) + (-9n + 12n + 9n) + (-6 - 9)
$$
$$
D_n = 12n - 15
$$

Now, find the sum of the first $$ n $$ first-differences:

$$
S_n = \sum_{k=1}^{n} D_k = \sum_{k=1}^{n} (12k - 15)
$$
$$
S_n = 12 \sum_{k=1}^{n} k - 15n
$$
$$
\sum_{k=1}^{n} k = \frac{n(n+1)}{2}
$$
$$
S_n = 12 \left( \frac{n(n+1)}{2} \right) - 15n = 6n(n+1) - 15n = 6n^2 + 6n - 15n = 6n^2 - 9n
$$
This contradicts the desired $$ S_n = 6n^2 + 3n $$. It seems there might be an error in the initial calculation.

Wait, let's re-examine the first difference calculation.

**Recalculating $$ D_n $$:**

Given $$ T_n = 6n^2 - 9n $$, then:

$$
D_n = T_n - T_{n-1} = (6n^2 - 9n) - [6(n-1)^2 - 9(n-1)]
$$
$$
= 6n^2 - 9n - [6(n^2 - 2n + 1) - 9n + 9]
$$
$$
= 6n^2 - 9n - 6n^2 + 12n - 6 + 9n - 9
$$
$$
= (6n^2 - 6n^2) + (-9n + 12n + 9n) + (-6 -9)
$$
$$
= 12n - 15
$$

Thus, $$ D_n = 12n - 15 $$.

Now, sum $$ D_n $$ from $$ k=1 $$ to $$ n $$:

$$
S_n = \sum_{k=1}^{n} (12k - 15) = 12 \sum_{k=1}^{n} k - 15n = 12 \cdot \frac{n(n+1)}{2} - 15n = 6n(n+1) - 15n = 6n^2 + 6n - 15n = 6n^2 - 9n
$$

But the problem states $$ S_{n} = 6n^{2} + 3n $$. There seems to be a discrepancy. However, considering the pattern provided:

Given the sequence: $$ -3 ; 6 ; 27 ; 60 ; \ldots $$

Compute first differences:

$$
6 - (-3) = 9 \
27 - 6 = 21 \
60 - 27 = 33 \
$$

Differences: 9, 21, 33,...

These differences themselves form an arithmetic sequence with a common difference of 12.

Thus, the sum of the first $$ n $$ differences:

$$
S_n = \sum_{k=1}^{n} (9 + 12(k-1)) = \sum_{k=1}^{n} (12k - 3) = 12 \cdot \frac{n(n+1)}{2} - 3n = 6n(n + 1) - 3n = 6n^2 + 6n - 3n = 6n^2 + 3n
$$

**Answer:** 
$$
S_n = 6n^2 + 3n
$$

---

### **2.2.4 How many consecutive first-differences were added to the first term of the quadratic number pattern to obtain a term in the quadratic number pattern that has a value of 21,060?**

**Solution:**

We need to find $$ n $$ such that when adding the first $$ n $$ first-differences to the first term, the resulting term is 21,060.

Given:

- First term ($$ T_1 $$) = -3
- Sum of first $$ n $$ first-differences ($$ S_n $$) = 6n^2 + 3n
- Desired term ($$ T $$) = 21,060

The relationship is:

$$
T = T_1 + S_n \
21,060 = -3 + 6n^2 + 3n \
6n^2 + 3n - 21,063 = 0 \
$$

Divide the entire equation by 3 to simplify:

$$
2n^2 + n - 7,021 = 0
$$

Now, solve the quadratic equation $$ 2n^2 + n - 7021 = 0 $$.

Use the quadratic formula:

$$
n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
$$

Where:
- $$ a = 2 $$
- $$ b = 1 $$
- $$ c = -7,021 $$

Calculate the discriminant ($$ \Delta $$):

$$
\Delta = 1^2 - 4(2)(-7021) = 1 + 56,168 = 56,169
$$

$$
\sqrt{56,169} = 237
$$

Now, find $$ n $$:

$$
n = \frac{-1 \pm 237}{4}
$$

We discard the negative solution as $$ n $$ must be positive:

$$
n = \frac{-1 + 237}{4} = \frac{236}{4} = 59
$$

**Answer:** 59 consecutive first-differences were added.

---

## **ION 3**

### **Problem**

**Given:** $$ \sum_{k=1}^{\infty} 4 \cdot 3^{2 - k} $$

1. **Prove that it is a convergent geometric series. Show all your calculations.**
2. **If $$ \sum_{k=p}^{\infty} 4 \cdot 3^{2 - k} = \frac{2}{9} $$, determine the value of $$ p $$.**

### **Solution:**

#### **1. Prove that it is a convergent geometric series.**

**Identify the series:**

$$
\sum_{k=1}^{\infty} 4 \cdot 3^{2 - k}
$$

Simplify the general term:

$$
a_k = 4 \cdot 3^{2 - k} = 4 \cdot 3^2 \cdot 3^{-k} = 4 \cdot 9 \cdot \left(\frac{1}{3}\right)^k = 36 \cdot \left(\frac{1}{3}\right)^k
$$

Recognize the geometric series form:

$$
a_k = ar^{k}
$$

Comparing:

$$
ar^{k} = 36 \cdot \left(\frac{1}{3}\right)^k
$$

Thus, the first term when $$ k = 1 $$:

$$
a = 36 \cdot \left(\frac{1}{3}\right)^1 = 36 \cdot \frac{1}{3} = 12
$$

The common ratio is:

$$
r = \frac{1}{3}
$$

**Criteria for convergence of a geometric series:**

A geometric series $$ \sum ar^{k} $$ converges if $$ |r| < 1 $$.

Here, $$ r = \frac{1}{3} $$, and $$ \left| \frac{1}{3} \right| < 1 $$.

**Conclusion:**

Since $$ |r| = \frac{1}{3} < 1 $$, the series is convergent.

**Answer:** The series is a convergent geometric series because the common ratio $$ r = \frac{1}{3} $$ satisfies $$ |r| < 1 $$.

#### **2. Determine the value of $$ p $$ if $$ \sum_{k=p}^{\infty} 4 \cdot 3^{2 - k} = \frac{2}{9} $$.**

**Solution:**

Given:

$$
\sum_{k=p}^{\infty} 4 \cdot 3^{2 - k} = \frac{2}{9}
$$

Simplify the series:

$$
\sum_{k=p}^{\infty} 4 \cdot 3^{2 - k} = \sum_{k=p}^{\infty} 36 \cdot \left(\frac{1}{3}\right)^k
$$

This is a geometric series with:

$$
a = 36 \cdot \left(\frac{1}{3}\right)^p \quad \text{and} \quad r = \frac{1}{3}
$$

The sum of an infinite geometric series starting from $$ k = p $$:

$$
S = \frac{a}{1 - r} = \frac{36 \cdot \left(\frac{1}{3}\right)^p}{1 - \frac{1}{3}} = \frac{36 \cdot \left(\frac{1}{3}\right)^p}{\frac{2}{3}} = 36 \cdot \left(\frac{1}{3}\right)^p \cdot \frac{3}{2} = 54 \cdot \left(\frac{1}{3}\right)^p
$$

Set this equal to $$ \frac{2}{9} $$:

$$
54 \cdot \left(\frac{1}{3}\right)^p = \frac{2}{9}
$$

Divide both sides by 54:

$$
\left(\frac{1}{3}\right)^p = \frac{2}{9} \div 54 = \frac{2}{9 \times 54} = \frac{2}{486} = \frac{1}{243}
$$

Express $$ \frac{1}{243} $$ as $$ \left(\frac{1}{3}\right)^5 $$ since $$ 3^5 = 243 $$:

$$
\left(\frac{1}{3}\right)^p = \left(\frac{1}{3}\right)^5
$$

Therefore:

$$
p = 5
$$

**Answer:** $$ p = 5 $$

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