In the quadratic sequence $$ 4 ; x ; y ;-11 ; \ldots $$, the first three terms of the first differences are $$ 2 p-4 ; p-3 $$ and $$ \frac{p}{2}-1 $$. Determine the: 4.1 . Value(s) of $$ p $$. 4.2 Second difference(s). 4.3 Values of $$ x $$ and $$ y $$. Given the quadratic sequence $$ 3 ; 5 ; 11 ; 21 ; \ldots $$ 5.1 Write down the value of the next term, if the pattern continues. 5.2 Determine the value of the $$ 48^{ ext {th }} $$ term. 5.3 Prove that the terms of this sequence will never be even numbers. 5.4 If all the values of this sequence are increased by 30 , determine th Gence. general term of the new sequence.

Question

In the quadratic sequence $$ 4 ; x ; y ;-11 ; \ldots $$, the first three terms of the first differences are $$ 2 p-4 ; p-3 $$ and $$ \frac{p}{2}-1 $$. Determine the: 4.1 . Value(s) of $$ p $$. 4.2 Second difference(s). 4.3 Values of $$ x $$ and $$ y $$. Given the quadratic sequence $$ 3 ; 5 ; 11 ; 21 ; \ldots $$ 5.1 Write down the value of the next term, if the pattern continues. 5.2 Determine the value of the $$ 48^{	ext {th }} $$ term. 5.3 Prove that the terms of this sequence will never be even numbers. 5.4 If all the values of this sequence are increased by 30 , determine th Gence. general term of the new sequence.

UpStudy AI · Accepted Answer

Let's solve each part of the problem step-by-step.

---

### **Problem 4**

**Given:**
- Quadratic sequence: $$ 4 ; x ; y ; -11 ; \ldots $$
- First three first differences: $$ 2p - 4 ; p - 3 ; \frac{p}{2} - 1 $$

#### **4.1. Value(s) of $$ p $$**

**Solution:**
For a quadratic sequence, the second differences are constant. Let's express the first differences based on the sequence terms:

1. **First difference:**
   $$
   x - 4 = 2p - 4 \quad \Rightarrow \quad x = 2p
   $$
   
2. **Second difference:**
   $$
   y - x = p - 3 \quad \Rightarrow \quad y = x + (p - 3) = 2p + p - 3 = 3p - 3
   $$
   
3. **Third difference:**
   $$
   -11 - y = \frac{p}{2} - 1 \quad \Rightarrow \quad -11 - (3p - 3) = \frac{p}{2} - 1
   $$
   $$
   -3p - 8 = \frac{p}{2} - 1
   $$
   $$
   -6p - 16 = p - 2 \quad \text{(Multiply both sides by 2)}
   $$
   $$
   -7p = 14 \quad \Rightarrow \quad p = -2
   $$

**Answer:**
$$
p = -2
$$

#### **4.2. Second Difference(s)**

**Solution:**
Using $$ p = -2 $$:

1. **First differences:**
   $$
   2p - 4 = 2(-2) - 4 = -8
   $$
   $$
   p - 3 = -2 - 3 = -5
   $$
   $$
   \frac{p}{2} - 1 = \frac{-2}{2} - 1 = -2
   $$
   
2. **Second differences:**
   $$
   (-5) - (-8) = 3
   $$
   $$
   (-2) - (-5) = 3
   $$

**Answer:**
$$
\text{Second difference is } 3
$$

#### **4.3. Values of $$ x $$ and $$ y $$**

**Solution:**
Using $$ p = -2 $$:

1. $$ x = 2p = 2(-2) = -4 $$
2. $$ y = 3p - 3 = 3(-2) - 3 = -9 $$

**Answer:**
$$
x = -4 \quad \text{and} \quad y = -9
$$

---

### **Problem 5**

**Given:**
- Quadratic sequence: $$ 3 ; 5 ; 11 ; 21 ; \ldots $$

#### **5.1. Next Term in the Sequence**

**Solution:**
1. **First differences:**
   $$
   5 - 3 = 2
   $$
   $$
   11 - 5 = 6
   $$
   $$
   21 - 11 = 10
   $$
   
2. **Second differences:**
   $$
   6 - 2 = 4
   $$
   $$
   10 - 6 = 4
   $$
   
3. **Next first difference:**
   $$
   10 + 4 = 14
   $$
   
4. **Next term:**
   $$
   21 + 14 = 35
   $$

**Answer:**
$$
\text{The next term is } 35
$$

#### **5.2. Value of the $$ 48^{\text{th}} $$ Term**

**Solution:**
1. **Find the general term $$ T(n) $$ for the sequence.**
   
   Assume $$ T(n) = an^2 + bn + c $$.
   
   Using the first three terms:
   $$
   \begin{cases}
   a(1)^2 + b(1) + c = 3 \
   a(2)^2 + b(2) + c = 5 \
   a(3)^2 + b(3) + c = 11 \
   \end{cases}
   $$
   
   Simplifying:
   $$
   \begin{cases}
   a + b + c = 3 \
   4a + 2b + c = 5 \
   9a + 3b + c = 11 \
   \end{cases}
   $$
   
   Subtract the first equation from the second:
   $$
   3a + b = 2
   $$
   
   Subtract the second equation from the third:
   $$
   5a + b = 6
   $$
   
   Subtract these two:
   $$
   2a = 4 \quad \Rightarrow \quad a = 2
   $$
   
   Then:
   $$
   3(2) + b = 2 \quad \Rightarrow \quad b = -4
   $$
   
   Finally:
   $$
   2 - 4 + c = 3 \quad \Rightarrow \quad c = 5
   $$
   
   Thus, the general term:
   $$
   T(n) = 2n^2 - 4n + 5
   $$
   
2. **Calculate the $$ 48^{\text{th}} $$ term:**
   $$
   T(48) = 2(48)^2 - 4(48) + 5 = 2(2304) - 192 + 5 = 4608 - 192 + 5 = 4421
   $$

**Answer:**
$$
\text{The } 48^{\text{th}} \text{ term is } 4421
$$

#### **5.3. Proof that All Terms are Odd Numbers**

**Solution:**
Given the general term:
$$
T(n) = 2n^2 - 4n + 5
$$
- $$ 2n^2 $$ is always even.
- $$ -4n $$ is always even.
- $$ 5 $$ is odd.

Thus:
$$
\text{Even} + \text{Even} + \text{Odd} = \text{Even} + \text{Odd} = \text{Odd}
$$

Therefore, every term $$ T(n) $$ is odd.

**Answer:**
$$
\text{All terms are odd because each term is an even number plus 5, making them always odd.}
$$

#### **5.4. General Term of the New Sequence (Each Term Increased by 30)**

**Solution:**
1. **Original general term:**
   $$
   T(n) = 2n^2 - 4n + 5
   $$
   
2. **New sequence with each term increased by 30:**
   $$
   T_{\text{new}}(n) = T(n) + 30 = 2n^2 - 4n + 5 + 30 = 2n^2 - 4n + 35
   $$

**Answer:**
$$
\text{The general term of the new sequence is } 2n^2 - 4n + 35
$$

---
**Problem 4:** 1. **Value of $$ p $$:** $$ p = -2 $$ 2. **Second Difference:** 3 3. **Values of $$ x $$ and $$ y $$:** $$ x = -4 $$, $$ y = -9 $$ **Problem 5:** 1. **Next Term:** 35 2. **48th Term:** 4421 3. **All Terms are Odd:** Yes, because each term is an even number plus 5, making them always odd. 4. **New General Term:** $$ 2n^2 - 4n + 35 $$

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