QUESTION 4 4.1 Given the following sequence of numbers: -5; 2; 9; ... 4.1.1 Identify the pattern (supply reason). (1) Mathematics Grade 10/P1 2 4.1.2 Write down $$ T_{4} $$. (1) 4.1.3 Determine the general term for the above sequence. (2) 4.1.4 Determine (using $$ T_{n} $$ ), the value of $$ 20^{ ext {th }} $$ term. (2) 4.1.5 Which term will be equal to 116 ? (2) 4.2 Consider the number pattern with a first difference equal to 6 . The sequence is $$ -8 ; x ; y ; 10 $$. 4.2.1 Determine by calculation the value of $$ x $$ and $$ y $$, (2) 4.2.2 Determine the general term. (2) 4.3 Consider the following pattern $$ \frac{5}{1} ; \frac{8}{4} ; \frac{11}{9} ;-; \ldots $$. 4.3.1 Write down the $$ 4^{ ext {th }} $$ term. (1) 4.3.2 Determine the general term for the sequence. (3)

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QUESTION 4 4.1 Given the following sequence of numbers: -5; 2; 9; ... 4.1.1 Identify the pattern (supply reason). (1) Mathematics Grade 10/P1 2 4.1.2 Write down $$ T_{4} $$. (1) 4.1.3 Determine the general term for the above sequence. (2) 4.1.4 Determine (using $$ T_{n} $$ ), the value of $$ 20^{	ext {th }} $$ term. (2) 4.1.5 Which term will be equal to 116 ? (2) 4.2 Consider the number pattern with a first difference equal to 6 . The sequence is $$ -8 ; x ; y ; 10 $$. 4.2.1 Determine by calculation the value of $$ x $$ and $$ y $$, (2) 4.2.2 Determine the general term. (2) 4.3 Consider the following pattern $$ \frac{5}{1} ; \frac{8}{4} ; \frac{11}{9} ;-; \ldots $$. 4.3.1 Write down the $$ 4^{	ext {th }} $$ term. (1) 4.3.2 Determine the general term for the sequence. (3)

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**4.1**

1. The sequence is $$-5,\,2,\,9,\,\dots$$. The difference between consecutive terms is  
   $$
   2 - (-5) = 7 \quad\text{and}\quad 9 - 2 = 7.
   $$
   Since the difference is constant, the pattern is an arithmetic progression with common difference $$ d = 7 $$.

2. The fourth term, using the formula  
   $$
   T_{n} = a + (n-1)d,
   $$
   is  
   $$
   T_{4} = -5 + 3(7) = -5 + 21 = 16.
   $$

3. The general term of an arithmetic sequence is given by  
   $$
   T_{n} = a + (n-1)d.
   $$
   Substituting $$ a = -5 $$ and $$ d = 7 $$ gives  
   $$
   T_{n} = -5 + 7(n-1) = 7n - 12.
   $$

4. The $$20^{\text{th}}$$ term is calculated as  
   $$
   T_{20} = -5 + 19(7) = -5 + 133 = 128.
   $$

5. To find which term is equal to 116, set the general term equal to 116:
   $$
   7n - 12 = 116.
   $$
   Solving for $$ n $$:
   $$
   7n = 116 + 12 = 128 \quad\Rightarrow\quad n = \frac{128}{7}.
   $$
   Since $$\frac{128}{7}$$ is not an integer, no term in the sequence is exactly 116.

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**4.2**

The given sequence is $$-8,\, x,\, y,\, 10$$, and the first difference is $$6$$.

1. Starting with the first term:
   $$
   x = -8 + 6 = -2,
   $$
   and
   $$
   y = -2 + 6 = 4.
   $$
   (The fourth term is $$4 + 6 = 10$$, which verifies the pattern.)

2. The general term for an arithmetic sequence is  
   $$
   T_{n} = a + (n-1)d.
   $$
   Here, $$ a = -8 $$ and $$ d = 6 $$, so
   $$
   T_{n} = -8 + (n-1)\cdot6 = 6n - 14.
   $$

---

**4.3**

The given pattern is
$$
\frac{5}{1} ;\, \frac{8}{4} ;\, \frac{11}{9} ;\, \ldots
$$

1. The numerators follow the pattern: $$5,\,8,\,11,\,\dots$$ with a common difference of $$3$$.  
   The denominators are perfect squares: $$1=1^2,\; 4=2^2,\; 9=3^2,\; \dots$$

The $$4^{\text{th}}$$ term has:
   - Numerator: $$11 + 3 = 14$$ (or computed as $$3(4)+2$$)
   - Denominator: $$4^2 = 16$$

Thus, the $$4^{\text{th}}$$ term is
   $$
   \frac{14}{16} = \frac{7}{8}.
   $$

2. The general term of the sequence:  
   For the numerators, note that when $$ n=1 $$: $$3(1)+2 = 5$$, when $$ n=2 $$: $$3(2)+2 = 8$$, etc.  
   For the denominators, the pattern is $$ n^2 $$.  
   Therefore, the general term is
   $$
   T_n = \frac{3n+2}{n^2}.
   $$
**4.1** 1. The sequence is arithmetic with a common difference of 7. 2. $$ T_{4} = 16 $$. 3. General term: $$ T_{n} = 7n - 12 $$. 4. $$ T_{20} = 128 $$. 5. No term equals 116. **4.2** 1. $$ x = -2 $$ and $$ y = 4 $$. 2. General term: $$ T_{n} = 6n - 14 $$. **4.3** 1. $$ T_{4} = \frac{7}{8} $$. 2. General term: $$ T_{n} = \frac{3n+2}{n^2} $$.

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