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 \( 20^{\text {th }} \) 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 \( 4^{\text {th }} \) term. (1) 4.3.2 Determine the general term for the sequence. (3)

**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. --- **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 \[ T_n = \frac{3n+2}{n^2}. \]