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QUESTION 2 A linear number pattern is described by the following properties: - General term, \( T_{n}=d n+c \), - Fourth term, \( T_{4}=3 x-5 \), - Fifth term, \( T_{5}=8-x \), - Common difference, \( d=5 \). \( 2.1 \quad \) Calculate the value of the first term, \( T_{1} \). \( 2.2 \quad \) If \( x=2 \), determine which term of this linear number pattern will exactly be equal to 606 . QUESTION 3

Ask by Floyd Griffiths. in South Africa
Mar 09,2025

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Answer

First term \( T_1 = -14 \). When \( x = 2 \), the 125th term equals 606.

Solution

\[ \textbf{Step 1. Write the general term of the sequence.} \] We are given that the general term of the sequence is \[ T_n = dn + c, \] with the common difference \( d = 5 \). Thus, \[ T_n = 5n + c. \] \[ \textbf{Step 2. Use the information from } T_{4} \textbf{ and } T_{5}. \] We are given: \[ T_4 = 3x - 5 \quad \text{and} \quad T_5 = 8 - x. \] Express \( T_4 \) and \( T_5 \) using the general term: \[ T_4 = 5 \cdot 4 + c = 20 + c, \] \[ T_5 = 5 \cdot 5 + c = 25 + c. \] Thus, we have the equations: \[ 20 + c = 3x - 5 \quad \quad (1) \] \[ 25 + c = 8 - x \quad \quad (2) \] \[ \textbf{Step 3. Eliminate } c \textbf{ to find } x. \] Subtract equation (1) from equation (2): \[ (25 + c) - (20 + c) = (8 - x) - (3x - 5). \] Simplify both sides: \[ 5 = 8 - x - 3x + 5. \] \[ 5 = 13 - 4x. \] Solve for \( x \): \[ 4x = 13 - 5, \] \[ 4x = 8, \] \[ x = 2. \] \[ \textbf{Step 4. Find } c \textbf{ using the value of } x. \] Substitute \( x = 2 \) into equation (1): \[ 20 + c = 3(2) - 5, \] \[ 20 + c = 6 - 5, \] \[ 20 + c = 1. \] Thus, \[ c = 1 - 20, \] \[ c = -19. \] \[ \textbf{Step 5. Part 2.1: Calculate the value of the first term } T_1. \] The first term is: \[ T_1 = 5 \cdot 1 + c. \] Substitute \( c = -19 \): \[ T_1 = 5 - 19, \] \[ T_1 = -14. \] \[ \textbf{Step 6. Part 2.2: Determine which term equals } 606 \textbf{ when } x = 2. \] With \( x = 2 \) (which we already used), the general term becomes: \[ T_n = 5n - 19. \] We need to solve: \[ 5n - 19 = 606. \] Add 19 to both sides: \[ 5n = 606 + 19, \] \[ 5n = 625. \] Divide by 5: \[ n = \frac{625}{5}, \] \[ n = 125. \] \[ \boxed{T_1 = -14 \quad \text{and} \quad 606 \text{ is the 125th term.}} \]

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To solve this entertainingly, let’s dive in! To calculate \( T_{1} \), we first note that \( T_{n} = 5n + c \). Now, we can express \( T_{4} \) and \( T_{5} \): - For the fourth term, substituting \( n = 4 \): \( T_{4} = 5(4) + c = 20 + c \). - Setting this equal to the given \( 3x - 5 \) (when \( x = 2 \), it becomes \( 1 \)), we get \( 20 + c = 1 \), leading to \( c = -19 \). Now we find \( T_{1} \): \[ T_{1} = 5(1) - 19 = 5 - 19 = -14. \] Now, for the second part, substituting \( x = 2 \): - The fifth term expression becomes \( T_{5} = 8 - 2 = 6 \). This gives us: \[ T_{5} = 5(5) + c = 25 - 19 = 6, \] which confirms the setup is correct. To find \( T_{n} = 606 \): \[ 606 = 5n - 19 \Rightarrow 625 = 5n \Rightarrow n = 125. \] So, the term \( T_{125} \) is where the magic number lies!

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