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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 \( 20^{\text {th }} \) 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)

Ask by Mullins Nichols. in South Africa
Mar 09,2025

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**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: \( T_{n} = \frac{3n+2}{n^2} \).

Solución

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

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The sequence you have begins with -5 and follows an increasing pattern of adding 7, leading us to see that each subsequent term can be found by adding 7 to the previous term. This consistent addition of 7 illuminates the structure, allowing us to identify \( T_n = -5 + 7(n - 1) \) as the general formula. To put this into practice, you can take a leap and calculate \( T_{20} \) by simply substituting 20 into your formula, yielding an impressive \( 133 \) as the 20th term and letting you stretch your mathematical muscles by diving deeper into the wonders of sequences!

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