8.1 Calculate $$ \sum_{r=1}^{13} 3 $$ 8.2 Given the arithmetic sequence: $$ 3 ; b ; 19 ; 27 ; \ldots $$ 8.2.1 Calculate the value of $$ b $$. 8.2.2 Determine the $$ n^{ ext {th }} $$ term of the sequence. 8.2.3 Calculate the value of the thirtieth term $$ \left(T_{30} ight) $$ 8.2.4 Calculate the sum of the first 30 terms of the sequence. The above sequence $$ 3 ; b ; 19 ; 27 ; \ldots $$ forms the first differences of a quadratio sequence. The first term of the quadratic sequence is 1 . 8.3.1 Determine the fourth term $$ \left(T_{4} ight) $$ of the quadratic sequence. 8.3.2 Determine the $$ n^{ ext {th }} $$ term of the quadratic sequence. 8.3.3 Calculate the value of $$ n $$ if $$ T_{n}-1=7700 $$

Question

8.1 Calculate $$ \sum_{r=1}^{13} 3 $$ 8.2 Given the arithmetic sequence: $$ 3 ; b ; 19 ; 27 ; \ldots $$ 8.2.1 Calculate the value of $$ b $$. 8.2.2 Determine the $$ n^{	ext {th }} $$ term of the sequence. 8.2.3 Calculate the value of the thirtieth term $$ \left(T_{30}ight) $$ 8.2.4 Calculate the sum of the first 30 terms of the sequence. The above sequence $$ 3 ; b ; 19 ; 27 ; \ldots $$ forms the first differences of a quadratio sequence. The first term of the quadratic sequence is 1 . 8.3.1 Determine the fourth term $$ \left(T_{4}ight) $$ of the quadratic sequence. 8.3.2 Determine the $$ n^{	ext {th }} $$ term of the quadratic sequence. 8.3.3 Calculate the value of $$ n $$ if $$ T_{n}-1=7700 $$

UpStudy AI · Accepted Answer

Certainly! Let's tackle each part of your question step by step.

---

### **8.1 Calculate $$ \sum_{r=1}^{13} 3 $$**

**Solution:**

The summation $$ \sum_{r=1}^{13} 3 $$ means adding the number 3 a total of 13 times.

$$
\sum_{r=1}^{13} 3 = 3 + 3 + \ldots + 3 \quad (\text{13 terms}) = 3 \times 13 = 39
$$

**Answer:**  
$$\sum_{r=1}^{13} 3 = 39$$

---

### **8.2 Given the Arithmetic Sequence: $$ 3 ; b ; 19 ; 27 ; \ldots $$**

Since it's an arithmetic sequence, the difference between consecutive terms is constant.

#### **8.2.1 Calculate the value of $$ b $$.**

**Solution:**

Let the common difference be $$ d $$.

$$
\begin{align*}
b &= 3 + d \quad \text{(Term 2)} \
19 &= b + d \quad \text{(Term 3)} \
\end{align*}
$$

Substitute $$ b = 3 + d $$ into the second equation:

$$
19 = (3 + d) + d \
19 = 3 + 2d \
2d = 16 \
d = 8
$$

Now, find $$ b $$:

$$
b = 3 + 8 = 11
$$

**Answer:**  
$$ b = 11 $$

---

#### **8.2.2 Determine the $$ n^{\text{th}} $$ term of the sequence.**

**Solution:**

The formula for the $$ n^{\text{th}} $$ term of an arithmetic sequence is:

$$
a_n = a_1 + (n - 1)d
$$

Given:
- $$ a_1 = 3 $$
- $$ d = 8 $$

Plugging in:

$$
a_n = 3 + (n - 1) \times 8 = 3 + 8n - 8 = 8n - 5
$$

**Answer:**  
$$ a_n = 8n - 5 $$

---

#### **8.2.3 Calculate the value of the thirtieth term $$ \left(T_{30}\right) $$.**

**Solution:**

Using the $$ n^{\text{th}} $$ term formula:

$$
a_{30} = 8 \times 30 - 5 = 240 - 5 = 235
$$

**Answer:**  
$$ T_{30} = 235 $$

---

#### **8.2.4 Calculate the sum of the first 30 terms of the sequence.**

**Solution:**

The sum of the first $$ n $$ terms of an arithmetic sequence is:

$$
S_n = \frac{n}{2} \times (a_1 + a_n)
$$

Given:
- $$ n = 30 $$
- $$ a_1 = 3 $$
- $$ a_{30} = 235 $$

Plugging in:

$$
S_{30} = \frac{30}{2} \times (3 + 235) = 15 \times 238 = 3,570
$$

**Answer:**  
Sum of the first 30 terms is **3,570**

---

### **8.3 The sequence $$ 3 ; 11 ; 19 ; 27 ; \ldots $$ forms the first differences of a quadratic sequence. The first term of the quadratic sequence is 1.**

#### **8.3.1 Determine the fourth term $$ \left(T_{4}\right) $$ of the quadratic sequence.**

**Solution:**

Given:
- First term of quadratic sequence, $$ Q_1 = 1 $$
- First differences: $$ d_1 = 3 $$, $$ d_2 = 11 $$, $$ d_3 = 19 $$, $$ d_4 = 27 $$, ...

Compute subsequent terms:

$$
\begin{align*}
Q_2 &= Q_1 + d_1 = 1 + 3 = 4 \
Q_3 &= Q_2 + d_2 = 4 + 11 = 15 \
Q_4 &= Q_3 + d_3 = 15 + 19 = 34 \
\end{align*}
$$

**Answer:**  
$$ T_4 = 34 $$

---

#### **8.3.2 Determine the $$ n^{\text{th}} $$ term of the quadratic sequence.**

**Solution:**

A quadratic sequence has the general form:

$$
Q_n = an^2 + bn + c
$$

Given the first four terms:

$$
\begin{align*}
Q_1 &= a(1)^2 + b(1) + c = a + b + c = 1 \
Q_2 &= a(2)^2 + b(2) + c = 4a + 2b + c = 4 \
Q_3 &= a(3)^2 + b(3) + c = 9a + 3b + c = 15 \
\end{align*}
$$

Subtract equations to eliminate $$ c $$:

$$
\begin{align*}
Q_2 - Q_1: &\quad 3a + b = 3 \quad \text{(1)} \
Q_3 - Q_2: &\quad 5a + b = 11 \quad \text{(2)} \
\end{align*}
$$

Subtract equation (1) from equation (2):

$$
(5a + b) - (3a + b) = 11 - 3 \
2a = 8 \
a = 4
$$

Plug $$ a = 4 $$ into equation (1):

$$
3(4) + b = 3 \
12 + b = 3 \
b = -9
$$

Now, find $$ c $$ using $$ Q_1 = 1 $$:

$$
4 + (-9) + c = 1 \
-5 + c = 1 \
c = 6
$$

Thus, the $$ n^{\text{th}} $$ term is:

$$
Q_n = 4n^2 - 9n + 6
$$

**Answer:**  
$$ Q_n = 4n^2 - 9n + 6 $$

---

#### **8.3.3 Calculate the value of $$ n $$ if $$ T_n - 1 = 7700 $$.**

**Solution:**

Given:

$$
T_n - 1 = 7700 \Rightarrow T_n = 7701
$$

Using the quadratic formula derived earlier:

$$
Q_n = 4n^2 - 9n + 6 = 7701
$$

Set up the equation:

$$
4n^2 - 9n + 6 - 7701 = 0 \
4n^2 - 9n - 7695 = 0
$$

Solve for $$ n $$ using the quadratic formula:

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

Where:
- $$ a = 4 $$
- $$ b = -9 $$
- $$ c = -7695 $$

Calculate the discriminant:

$$
\Delta = (-9)^2 - 4(4)(-7695) = 81 + 123,120 = 123,201
$$

$$
\sqrt{123,201} = 351
$$

Now, find $$ n $$:

$$
n = \frac{9 \pm 351}{8}
$$

Discard the negative solution as $$ n $$ must be positive:

$$
n = \frac{9 + 351}{8} = \frac{360}{8} = 45
$$

**Answer:**  
$$ n = 45 $$

---
**Summary of Solutions:** 1. **8.1:** $$\sum_{r=1}^{13} 3 = 39$$ 2. **8.2.1:** $$ b = 11 $$ 3. **8.2.2:** $$ a_n = 8n - 5 $$ 4. **8.2.3:** $$ T_{30} = 235 $$ 5. **8.2.4:** Sum of first 30 terms = **3,570** 6. **8.3.1:** $$ T_4 = 34 $$ 7. **8.3.2:** $$ Q_n = 4n^2 - 9n + 6 $$ 8. **8.3.3:** $$ n = 45 $$

Real Tutor Solution

Answer

Solution

Mind Expander

Related Questions

Latest Algebra Questions