3 The sum of the second and sixth terms of an arithmetic sequence is 4 .
The third term is 24 more than the term.
3.1 Determine the first three terms in the sequence.
3.2 Determine a formula for the term of the sequence.
3.3 Determine the term in the sequence.
3.4 Is -72 a term in the sequence? Justify your answer.
4 Which term in the sequence has a value of 8 ?
5 Consider the sequence .
5.1 Determine a formula for the term of the sequence.
5.2 Show that , for

Ask by Reeves King. in South Africa

Mar 18,2025

Question

3 The sum of the second and sixth terms of an arithmetic sequence is 4 .
The third term is 24 more than the term.
3.1 Determine the first three terms in the sequence.
3.2 Determine a formula for the term of the sequence.
3.3 Determine the term in the sequence.
3.4 Is -72 a term in the sequence? Justify your answer.
4 Which term in the sequence has a value of 8 ?
5 Consider the sequence .
5.1 Determine a formula for the term of the sequence.
5.2 Show that , for

Ask by Reeves King. in South Africa

Mar 18,2025

UpStudy AI · Accepted Answer

**3.1 Determine the first three terms in the sequence.**

Let the first term be $$ a $$ and the common difference be $$ d $$.

- The second term is $$ a+d $$ and the sixth term is $$ a+5d $$. The sum is given by  
  $$
  (a+d)+(a+5d)=2a+6d=4.
  $$
  Dividing by 2, we have  
  $$
  a+3d=2. \tag{1}
  $$

- The third term is $$ a+2d $$ and the $$11^{\mathrm{th}}$$ term is $$ a+10d $$. It is given that  
  $$
  a+2d = (a+10d)+24.
  $$
  Simplify to obtain  
  $$
  a+2d-a-10d=24 \quad\Longrightarrow\quad -8d=24.
  $$
  Thus,  
  $$
  d=-3.
  $$

- Substitute $$ d=-3 $$ into equation (1):  
  $$
  a+3(-3)=2 \quad\Longrightarrow\quad a-9=2 \quad\Longrightarrow\quad a=11.
  $$

Therefore, the first three terms are:  
$$
\begin{array}{l}
T_1 = 11,\[1mm]
T_2 = 11+(-3)=8,\[1mm]
T_3 = 11+2(-3)=5.
\end{array}
$$

---

**3.2 Determine a formula for the $$ n^{\text{th}} $$ term of the sequence.**

The formula for the $$ n^{\text{th}} $$ term in an arithmetic sequence is  
$$
T_n = a + (n-1)d.
$$
Substitute $$ a=11 $$ and $$ d=-3 $$:  
$$
T_n = 11 + (n-1)(-3)=11-3n+3=14-3n.
$$

---

**3.3 Determine the $$ 17^{\text{th}} $$ term in the sequence.**

Using the formula $$ T_n = 14-3n $$ for $$ n=17 $$:  
$$
T_{17}=14-3(17)=14-51=-37.
$$

---

**3.4 Is $$-72$$ a term in the sequence? Justify your answer.**

To determine whether $$-72$$ is a term, we set  
$$
14-3n=-72.
$$
Solve for $$ n $$:
$$
14-3n=-72 \quad\Longrightarrow\quad -3n=-72-14=-86 \quad\Longrightarrow\quad n=\frac{86}{3}.
$$
Since $$ \frac{86}{3} $$ is not an integer, $$-72$$ is not a term in the sequence.

---

**4. Which term in the sequence $$ -2; -\frac{5}{3}; -\frac{4}{3}; -1; \ldots $$ has a value of 8?**

Let the first term be $$ a_1=-2 $$ and let the common difference be $$ d $$.

- The difference between the second and first terms is  
  $$
  -\frac{5}{3}-(-2)=-\frac{5}{3}+2=\frac{-5+6}{3}=\frac{1}{3}.
  $$
  Hence, $$ d=\frac{1}{3} $$.

- The $$ n^{\text{th}} $$ term is given by  
  $$
  T_n = a_1 + (n-1)d = -2+\frac{n-1}{3}.
  $$
  Writing with a common denominator:  
  $$
  T_n=\frac{-6+n-1}{3}=\frac{n-7}{3}.
  $$

- Set $$ T_n=8 $$:  
  $$
  \frac{n-7}{3}=8 \quad\Longrightarrow\quad n-7=24 \quad\Longrightarrow\quad n=31.
  $$

Thus, the $$ 31^{\text{st}} $$ term of the sequence is 8.

---

**5.1 Determine a formula for the $$ n^{\text{th}} $$ term of the sequence $$ 13; 11; 9; 7; 5; \ldots $$.**

Let the first term be $$ a=13 $$ and the common difference be $$ d $$.

- The second term is 11, so  
  $$
  d=11-13=-2.
  $$

The $$ n^{\text{th}} $$ term is given by  
$$
T_n = a + (n-1)d = 13 + (n-1)(-2)=13-2(n-1).
$$
Simplify the expression:  
$$
T_n = 13-2n+2=15-2n.
$$

---

**5.2 Show that $$ T_n+T_{15-n}=0 $$, for $$ n\in \mathbb{N},\, n<15 $$.**

Using the formula $$ T_n=15-2n $$:

- The $$ n^{\text{th}} $$ term is  
  $$
  T_n=15-2n.
  $$

- The $$(15-n)^{\text{th}}$$ term is  
  $$
  T_{15-n}=15-2(15-n)=15-30+2n=2n-15.
  $$

Now, add $$ T_n $$ and $$ T_{15-n} $$:
$$
T_n+T_{15-n}=(15-2n)+(2n-15)=0.
$$

Thus, it is shown that  
$$
T_n+T_{15-n}=0
$$
for all natural numbers $$ n $$ such that $$ n<15 $$.
**3.1 First Three Terms:** - $$ T_1 = 11 $$ - $$ T_2 = 8 $$ - $$ T_3 = 5 $$ **3.2 $$ n^{\text{th}} $$ Term Formula:** $$ T_n = 14 - 3n $$ **3.3 $$ 17^{\text{th}} $$ Term:** $$ T_{17} = -37 $$ **3.4 Is $$-72$$ a Term?** No, $$-72$$ is not a term in the sequence. **4. Term with Value 8:** The $$ 31^{\text{st}} $$ term is 8. **5.1 $$ n^{\text{th}} $$ Term Formula:** $$ T_n = 15 - 2n $$ **5.2 Verification:** $$ T_n + T_{15-n} = 0 $$ for all $$ n < 15 $$.

Upstudy AI Solution

Answer

Solution

Bonus Knowledge

Related Questions

Latest Algebra Questions