QUESTION 2 2.1 The following sequence is a combination of arithmetic and a geometric sequence: $$ 2 ; 3 ; 4 ; 6 ; 6 ; 12 ; \ldots $$ 2.1.1 Write down the next two terms of the sequence. 2.1.2 Calculate $$ \mathrm{T}_{30}-\mathrm{T}_{31} $$. 2.1.3 Write the geometric sequence in sigma notation. 2.2 The middle term of an arithmetic sequence of 31 terms is 50 . (3) Determine the sum of the series.

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$$ \textbf{2.1.1} $$  
Separate the sequence into two parts: the odd‐numbered terms and the even‐numbered terms.

• The odd terms are:  
$$
T_1=2,\quad T_3=4,\quad T_5=6,\ldots  
$$  
This is an arithmetic sequence with first term $$2$$ and common difference $$2$$. Thus, the next odd term is  
$$
T_7=6+2=8.
$$

• The even terms are:  
$$
T_2=3,\quad T_4=6,\quad T_6=12,\ldots  
$$  
This is a geometric sequence with first term $$3$$ and common ratio $$2$$. Hence, the next even term is  
$$
T_8=12\cdot 2=24.
$$

So, the next two terms of the sequence are:  
$$
8\text{ and }24.
$$

---

$$ \textbf{2.1.2} $$  
Identify the type of each term:

• $$ T_{30} $$ is an even-numbered term. In the even sequence, the general term is given by  
$$
T_{2k}=3\cdot 2^{\,k-1},
$$  
where $$ k $$ is the position within the even terms. For $$ T_{30} $$, we have  
$$
2k=30\Longrightarrow k=15,
$$  
so  
$$
T_{30}=3\cdot 2^{14}.
$$

• $$ T_{31} $$ is an odd-numbered term. The odd sequence is arithmetic with general term  
$$
T_{2k-1}=2+2(k-1).
$$  
For $$ T_{31} $$, the index satisfies  
$$
2k-1=31\Longrightarrow k=16,
$$  
thus  
$$
T_{31}=2+2\cdot15=32.
$$

Now compute  
$$
T_{30}-T_{31}=3\cdot 2^{14}-32.
$$

This is the required expression; if desired, one may note that  
$$
3\cdot 2^{14}=49152 \quad \text{and} \quad 49152-32=49120.
$$

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$$ \textbf{2.1.3} $$  
The geometric sequence (even-numbered terms) is  
$$
3,\, 6,\, 12,\, \ldots
$$  
It can be written in sigma notation for the sum of its first $$ n $$ terms as  
$$
\sum_{k=1}^{n} 3\cdot 2^{\,k-1}.
$$

---

$$ \textbf{2.2} $$  
For an arithmetic sequence of 31 terms, the middle term is the 16th term. Since the middle term is the average of all terms, the sum $$ S $$ of the series is given by  
$$
S= \text{(number of terms)}\times \text{(middle term)}.
$$

Thus,  
$$
S=31\times 50=1550.
$$
**2.1.1** The next two terms of the sequence are **8** and **24**. **2.1.2** Calculate $$ T_{30} - T_{31} = 3 \times 2^{14} - 32 $$. **2.1.3** The geometric sequence can be written in sigma notation as $$ \sum_{k=1}^{n} 3 \times 2^{k-1} $$. **2.2** The sum of the series is **1550**.

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